Topology Explained – The Bestiary: The Topologist’s Sine Curve

As a boy, I loved going to zoos. As an adult, I still love going to zoos; though with additional mental capacities brought on by age, such as empathy and object permanence, I do feel quite bad about keeping animals caged for our amusement (except, of course, for the drop bears). None-the-less, I have always wanted to visit a bestiary; a place where animals too abhorrent for public eye are kept, often with some kind of Christian moral associated with their confinement. Sadly, the closest I have ever come to actually visiting such a thing was the time I went to a local state fair where a carney got stuck on “the Zipper”:

Poor man was stuck in the thing for hours, like some kind of neon-tubed, jubilant iron maiden.

Photo Credit: By Juniortresha – Own work, CC BY-SA 3.0,

That, alas, is a story for another post. As a mathematician, however, I have come very close to finding such a place. Invaluable to the budding math student is the example: a wholesome expos√© which dutifully lays out the work, techniques, and thought process behind the material the student is currently learning. The antithesis to the example is, of course, the counter example, which I always found odd considering a counter example is really always an example in-of-itself, it’s just an example where things go wrong. To the mathematician, however, examples form a “zoo”, of sorts: they show you what happens when you play by the rules, generally are pleasant to work with, and sometimes a particularly stubborn example requires a bag of salted peanuts to coerce into doing what you want.

“And if you’ll look to your left, you’ll see the Gaussian Integral sulking in the corner because we ran out of sardines last night.”

Photo Credit: By Photo © Samuel Blanc, CC BY-SA 3.0,

Counter examples, however, well counter examples show you all of the evil in the world, they show you what happens if you stray off the beaten path, and they show you what will happen if you remove yourself from the protection of God’s light. As a bit of a contrarian/sadist myself, I have always found room in the cavity of my chest where my heart should have been for a good counter example; it does the portion of my soul I haven’t bargained away good to know that there’s something else out there as inconvenient and dissatisfying as me.

Chief among counter examples in mathematics is the topologist’s sine curve; an example of a topological space which is connected, but not path connected. While I will go into the details of what this means temporarily, effectively this means that it is a space that you cannot “break apart”; you can’t somehow cleanly split the space into halves, or even two disproportionate pieces. However, the space fails to be path connected, which roughly means that you cannot reach one point from another. While doing some research to refresh my memory about the damn thing, I ran across the following article by Evelyn Lamb which has, I think, the most succinct description of a connected, non-path connected space: “you can see the finish line, but you can’t get there from here“. Let’s get into it.


Grammarness be damned; I’m calling this section connectedness instead of “connected”. If you’re comfortable with a few basic results on a connected space, you can feel free to skip this section.

Definition: Let X be a topological space. We say that X is disconnected provided there are open subspaces U,V\subseteq X such that:

  1. U\cup V=X, so that U,V comprise all of X.
  2. U\cap V=\varnothing, so that U,V have no overlap.
  3. U,V\ne\varnothing, so that we can rule out stupid trivial cases.

Such a pair of open subspaces U,V of X is called a separation of X (note: this is non-standard terminology, just my own shorthand). Should X be not disconnected, we say that X is connected. Equivalently, X is connected if for any open subspaces U,V\subseteq X either: one of U,V is empty, or U\cup V\ne X, or U\cap V\ne\varnothing. Note that by taking complements, we can easily restate the definition of connectedness/disconnectedness in terms of closed subspaces, should we so choose.

There is, of course, an alternate way of describing connectedness.

Proposition: Let X be a space. Then X is connected if and only if the only clopen subspaces of X are \varnothing and X itself.

Proof: Suppose first that X,\varnothing are the only clopen sets. Suppose, however, that X were disconnected; by definition this implies that there are open U,V\subseteq X such that: U,V\ne\varnothing, U\cup V=X and U\cap V=\varnothing. Note that as U\cap V=\varnothing and U\cup V=X, this implies that U^C=V and likewise that V^C=U. However, as V is open, this implies that V^C=U is closed. Thus as U is open, U is then open and closed, i.e. clopen. But by assumption U\ne \varnothing and as V=U^C\ne\varnothing this implies that U\ne X either. Thus U is a clopen set which is neither \varnothing nor X, contradicting our assumption that these are the only clopen sets. Thus no such U,V can exist, hence X must be connected.

The converse is proved by arguing much in the same way as the above; suppose X is connected, but there is some U other than \varnothing,X which is clopen. Then U,U^C will form a separation of X, a contradiction.


Proposition: Any interval of \mathbb{R} (including the entire line itself) is connected.

Proof: We simply prove that a closed interval of the form [a,b] is connected; the proof for open/half-open/infinite rays is identical:

Yep, that’s an interval.

Suppose then that [a,b] is disconnected; let U,V be a separation of [a,b]. Now neither U,V can be empty, so suppose c\in U and d\in V; we may assume that c<d by relabeling U,V if necessary:

I have to ask that, for the sake of the illusion, ignore that – as drawn – neither U,V are open.

Now, as c,d\in[a,b] and c<d, this certainly implies that [c,d]\subseteq [a,b]. Moreover, as U\cup V=[a,b], it follows that [c,d]\subseteq U\cup V. As such, every element of [c,d] is an element of either U or V, and we also know that c\in U and d\in V so that the interval [c,d] is not contained entirely within either U,V:

I’m running out of colors

Let U'=[c,d]\setminus V=[c,d]\cap V^C=[c,d]\cap U. The claim is that U' is compact; this follows as it is a closed subspace of a Hausdorff space ([c,d] is Hausdorff as a subspace of \mathbb{R}, and V^C is closed as the complement of an open space, thus U' is a closed subspace of [c,d]). But any non-empty compact subspace of \mathbb{R} has a maximal element; this is a standard result of real analysis. As U' is compact and non-empty (as c\in [c,d] and c\notin V), U' then has a maximal element, say x. As x\notin V, we must have that x\in U (as U\cup V\supseteq[c,d]), and hence that x<d:

It seems so obvious when you draw the damn picture.

Similarly, let V'=[x,d]\setminus U=[x,d]\cap U^C. Analogous arguments (e.g. V' is non-empty, for we know that x<d and d\in V so that d\in[x,d]\setminus U) show that V' must have a minimal element, say y. Then certainly x\le y; as x\in U and y\notin U, we must also have that x<y.

Finally, the punch line. As x<y, there must be some z such that x<z<y. By construction, y is the smallest element of [x,d]\setminus U, so if z<y we must have z\notin V (else z\notin U and hence z\in[x,d]\setminus U and z<y, a contradiction). However, as c\le x<z<y\le d, we must have that z\in[c,d]. As z\in[c,d] and z\notin U, we must have z\in U'=[c,d]\setminus V. But then x,z\in U' with x<z, contradicting the fact the x is the maximal element of U'. As such, we see that no such U,V can exist, so that the interval [a,b] is connected.


In fact, the above can be strengthened considerably. It turns out that not only are intervals connected subspaces of \mathbb{R}, they’re the only connected subspaces of \mathbb{R}.

Proposition: Let X\subseteq \mathbb{R}. If X is connected (as a subspace of \mathbb{R}), then X is an interval.

Proof: Suppose that X is connected yet X is not an interval. First note that every singleton is an interval, so we may assume that X has at least two distinct points. Then there must be points a,b\in X such that a<b and yet there is some a<c<b such that c\notin X (for if no such a,b exist, then X is an interval). Then consider X^+=(c,+\infty)\cap X and X^-=(-\infty,c)\cap X. Clearly X^+,X^- are open subsets of X, as they are open intervals intersected with X. Clearly X^+\cap X^-=\varnothing, as otherwise we would have some x\in X^-\cap X^+, which would imply we would have some x\in(-\infty,c)\cap(c,+\infty), i.e. we would have some number x such that x<c and x>c. Finally, since c\notin X, we also then must have that X^-\cup X^+=X. Thus X^-,X^+ form a separation of X so that X is not connected, a contradiction. Thus X must be an interval.


Proposition: Let X,Y be spaces, and let X\xrightarrow{f}Y be a continuous map. If X is connected, then f(X) is a connected subspace of Y.

Proof: Note that some authors may write \text{Im}(f) in place of f(X). In any case, suppose that X is connected, but f(X) were not connected; say that W,Z is a separation of f(X). That is, W,Z are open subspaces of f(X), W\cup Z=f(X), and W\cap Z=\varnothing. As W,Z are open subspaces of f(X), by definition of the subspace topology this implies that there are open W',Z'\subseteq Y such that W=f(X)\cap W' and Z=f(X)\cap Z. Set U=f^{-1}(W') and V=f^{-1}(Z'); the claim is that U,V then form a separation of X, contradicting the connectedness of X.

Then first note that U,V are open in X as they are both the pullbacks of open sets by a continuous map. Now W,Z are both non-empty; say w\in W and z\in Z. By definition, this implies that w\in f(X)\cap W' and z\in f(X)\cap Z'. As such, both w,z\in f(X) so that there are x,y\in X such that f(x)=w and f(y)=z. But then, as w\in W' and z\in Z', this implies that x\in f^{-1}(W') and y\in f^{-1}(Z'), so that x\in U and y\in V. Thus neither of U,V are empty. Suppose, however, that U\vap V\ne\varnothing; say x\in U\cap V. Then certainly f(x)\in f(X). Moreover, this implies that x\in f^{-1}(W')\cap f^{-1}(Z') so that f(x)\in W'\cap Z'. But this in turn implies that f(x)\in f(X)\cap W' and f(x)\in f(X)\cap Z', i.e. f(x)\in W and f(x)\in Z, i.e. f(x)\in W\cap Z, but this contradicts the fact that W\cap Z=\varnothing as W,Z are a separation. Thus we must have that U\cap V=\varnothing. Finally, take any x\in X. Certainly f(x)\in f(X); as f(x)=W\cup Z we must have that either f(x)\in W or f(x)\in Z. If f(x)\in W, then f(x)\in W', and hence x\in f^{-1}(W')=U. Likewise, if f(x)\in Z, then f(x)\in Z', and hence x\in f^{-1}(Z')=V. Thus we must have that x\in U\cup V, so that X\subseteq U\cup V. As the other containment follows by construction, we see that X=U\cup V.

Thus we’ve shown that U,V are non-empty open subsets of X such that U\cap V=\varnothing and U\cup V=X, so that U,V form a separation of X. By assumption, X is connected so no such separation can exist. Thus no such separation of f(X) can exist, so f(X) is connected.


These are the biggies when it comes to connectedness. Something perhaps surprising is that for such a simple criterion, it is typically rather difficult to prove if a given space is connected or not. Most often, the prior proposition is the main tool that gets used in doing so; one keeps a small arsenal of spaces that they know, or are commonly known to be, connected, then try and envision unknown spaces as the continuous images of these spaces. Next, we look at a similar idea, path connectedness.

Path Connectedness

To be blunt, path connectedness is the inherent ability of a space to “continuously” join two points. Most often, it is thought about as the ability to trace a “path” between two points in a space, without ever leaving the space to do so.

Definition: Let X be a space. A path in X is a continuous map [0,1]\xrightarrow{\gamma}X. Given a path \gamma in X, \gamma(0)\in X is called the starting point of the path \gamma, and \gamma(1)\in X is called the ending point of the path \gamma. X is called path connected if for any points x,y\in X, there is a path \gamma in X with x as its starting point and y as its end point.

Without going into too much detail, paths function rather nicely and intuitively. For example; should you have a path \gamma in a space X starting at x and ending at y, then you also have a path in X starting at y and ending at x, by following \gamma “backwards”. If you have a path \gamma from x to y and you have a path \eta from y to z, then you have a path from x to z by first following \gamma, then following \eta. We could go on all day proving nice things about paths, but instead let’s focus on the results that are important.

Proposition: Fix any n\in\mathbb{N}. Any convex subspace of \mathbb{R}^n is path connected. In particular, intervals are path connected.

Proof: Recall that if O\subseteq \mathbb{R}^n is convex, then this implies that for any \mathbf{x},\mathbf{y}\in O, the line \left\{(\mathbf{y}-\mathbf{x})t+\mathbf{x}:t\in[0,1]\right\} is also contained in O. But this effectively proves the proposition outright. Let O be an convex subspace of \mathbb{R}^n, and take any \mathbf{x},\mathbf{y}\in O. Define [0,1]\xrightarrow{\gamma}O by \gamma(t)=(\mathbf{y}-\mathbf{x})t+\mathbf{x}; note that we can indeed set the codomain of \gamma to be O as O is convex and contains all of these possible values. However, it is easy to check (by writing \mathbf{x}=(x_1,x_2,\ldots,x_n) and then writing \gamma in terms of component functions) that \gamma is in fact continuous, \gamma(0)=\mathbf{x}, and \gamma(1)=\mathbf{y}, and hence that \gamma is a path in O from \mathbf{x} to \mathbf{y}; as \mathbf{x},\mathbf{y}\in O were generic, this implies that O is path connected.


Proposition: Let X,Y be spaces, and X\xrightarrow{f}Y be a continuous function. If X is path connected, then f(X) is path connected.

Proof: Assume that X is path connected. Take any z,w\in f(X); by definition, there are then x,y\in X such that f(x)=z and f(y)=w. As X is path connected, there must be a path \gamma from x to y, i.e. a continuous map [0,1]\xrightarrow{\gamma}X such that \gamma(0)=x and \gamma(1)=y. Then certainly [0,1]\xrightarrow{f\circ\gamma}f(X) is a continuous function, and \big(f\circ\gamma\big)(0)=f\big(\gamma(0)\big)=f(x)=z and \big(f\circ\gamma\big)(1)=f\big(\gamma(1)\big)=f(y)=w so f\circ\gamma is a path in f(X) from z to w. As z,w\in f(X) were arbitrary, this implies that f(X) is path connected.


Theorem: Let X be a space. If X is path connected, then X is connected.

Proof: Assume the contrary, i.e. assume that X is path connected but not connected. As X is not connected, let U,V be a separation of X. As neither U,V are empty, let x\in U and y\in V. As X is path connected, let \gamma be a path in X from x to y. Now simply consider A=\gamma^{-1}(V) and B=\gamma^{-1}(U) as subsets of [0,1]. Clearly they are open as \gamma is continuous and U,V are open subsets of X. Moreover, they must be non-empty as 0\in \gamma^{-1}(U), since \gamma(0)=x\in U, and likewise 1\in \gamma^{-1}(V). Additionally, we must have that A\cap B=\varnothing. For suppose otherwise, i.e. suppose t\in A\cap B. Then t\in \gamma^{-1}(U) and t\in\gamma^{-1}(V) so that \gamma(t)\in U\cap V, contradicting the fact that U\cap V=\varnothing. Finally, we must have that A\cup B=[0,1]. For take any t\in[0,1]; then \gamma(t)\in X=U\cup V. Thus either \gamma(t)\in U or \gamma(t)\in V so that t\in\gamma^{-1}(U) or t\in\gamma^{-1}(V); in either case we ultimately then get that t\in A\cup B. Thus we have found non-empty open subsets A,B of [0,1] such that A\cap B=\varnothing and A\cup B=[0,1], so that A,B form a separation of [0,1]. This, however, contradicts the fact that [0,1] is connected, as we have previously proven. Thus no such separation U,V can exist, so X also must be connected.


The above theorem is essentially our motivation for this entire rant. We see that when it comes to topology, \text{path connected}\Rightarrow \text{connected}. But all mathematicians, when faced with an implication, must immediately try and determine if the converse is true. Here, it seems like it should be true; on the face of it, there is no immediate reason why a connected space should not be path connected. Moreover, most of the common examples of connected spaces are path connected, so anecdotally it also appears as though it should be true. Yet it is not true, and nearly every single student who takes a topology course gets tricked into thinking so by their instructors, simply so the instructor can pull back the curtain and show them the disgusting beast that is the topologist’s sine curve; the prototypical connected, not path connected space. We’ll go through the construction of the sine curve and prove that it has the appropriate properties. Then we’ll look more closely at what exactly makes the topologist’s sine curve behave so badly.

Setup and Construction

Set A=\left\{\left(t,\sin\left(\frac{1}{t}\right)\right):t\in(0,1]\right\}. Note then that A is precisely the graph of (0,1]\xrightarrow{f}\mathbb{R} with f(t)=\sin\left(\frac{1}{t}\right). Now let B=\left\{(0,t):t\in[-1,1]\right\}. Then B is also the graph of [-1,1]\xrightarrow{g}\mathbb{R} given by g(t)=t. Now let \Delta=A\cup B, and equip \Delta with the subspace topology; \Delta is the topologist’s sine curve, and it looks something like this:

Note that in actuality, there is no “gap” between A and B in the “real” picture; B butts up against A and the more rapidly that the sine wave oscillates as it gets closer to the line x=0, the hard it becomes to differentiate B from the tail end of A. Thus the “gap” in the picture is just for style, not to be a representative of anything.

Note that not all authors define the \Delta in this way. Many do the following: let \nabla=A\cup\left\{(0,0)\right\}. Basically you can think of this as just being A together with a single point, like so:

B just got out of the pool.

There is actually good reason for this, namely that \nabla=\overline{A}; we’ll look back at this later. However, I prefer the definition I’ve chose for artistic reasons, but it turns out that these two spaces are, more-or-less, the same. Now let’s prove the things we need about \Delta and \nabla.

\Delta is Connected

Lemma: A,B are both connected.

Proof: Recall that A is the graph of (0,1]\xrightarrow{f}\mathbb{R} given by f(t)=\sin\left(\frac{1}{t}\right). But we can also visualize this as the image of the function (0,1]\xrightarrow{F}\mathbb{R}^2 given by F(t)=\left(t,\sin\left(\frac{1}{t}\right)\right), and F is continuous (the verification of which, I leave to you). Similarly, we can realize B as the image of [-1,1]\xrightarrow{G}\mathbb{R}^2 given by G(t)=(0,t), which is also continuous. However, (0,1],[-1,1] are both connected as they are both intervals, but this implies that A,B are both connected as the images of connected spaces under continuous maps.


Theorem: \Delta is connected.

Proof: Suppose otherwise, i.e. suppose that U,V is a separation of \Delta, so that U,V are both open, non-empty, U\cap V=\varnothing, and U\cup V=\Delta. Now as U\cup V=\Delta and (0,0)\in\Delta we must have that (0,0) is in either U or V; without loss of generality, assume that (0,0)\in U. As we know that (0,0)\in B and (0,0)\in U, this implies that B\cap U\ne\varnothing. Now, however, we have two cases to consider.

Case 1: B\not\subseteq U. This implies that there is some x\in B such that x\notin U. But since B\subseteq \Delta and \Delta=U\cup V, this implies that x\in V, so that B\cap V\ne\varnothing. Now let C=B\cap U and D=B\cap V. Then we know that C,D are open, non-empty subspaces of D (open as U,V are open in \Delta, and non-empty as we just showed (0,0)\in B\cap U and x\in B\cap V). The claim is that C,D then form a separation of B. Note that C\cap D=\varnothing, for if C\cap D\ne\varnothing, this implies that there is some y\in C\cap D=\big(B\cap U\big)\cap\big(B\cap V\big)=B\cap\big(U\cap V\big), which implies that y\in U\cap V, contradicting the fact that U\cap V=\varnothing. However, we also must have that C\cup D=B. This follows, for

    \[C\cup D=\big(B\cap U\big)\cup\big(B\cap V\big)=(U\cup V)\cap B=\Delta\cap B=B\]

Thus, as claimed, C,D form a separation of B. This, however, is a contradiction as we proved in the preceding lemma that B is connected.

Case 2: B\subseteq U. In this case, we basically repeat the same argument, but with A. But there is a small subtle catch that takes some explaining before we really begin, namely we need to show that A\cap U,A\cap V\ne\varnothing. Now if \Delta=A\cup B, \Delta=U\cup V, U\cap V=\varnothing, and B\subseteq U, it follows that we must have that V\subseteq A; as V\ne\varnothing this certainly implies that A\cap V\ne\varnothing. Thus we really only need to show that A\cap U\ne\varnothing. To do this, note that (0,0)\in U, and U is an open subset of \Delta. However, \Delta is a subspace of \mathbb{R}^2, meaning that U=W\cap \Delta for some open W\subseteq \mathbb{R}^2. But then W must contain some open ball B_{\varepsilon about (0,0), meaning that U must also contain the intersection of this ball with \Delta. But for any \epsilon>0, the open ball of radius \epsilon about (0,0) will intersect A, as seen below:

Mmm, yes, quite.

However, this is topology explained, not topology, so given a ball about (0,0) of radius \varepsilon>0, take any N\in\mathbb{N} greater than \frac{1}{2\pi\varepsilon}, and consider the point \left(\frac{1}{2\pi N},0\right). As N>\frac{1}{2\pi\varepsilon}, this in turn implies that \varepsilon>\frac{1}{2\pi N}. But then,

    \begin{align*}\left|\left(\frac{1}{2\pi N},0\right)-\left(0,0\right)\right| & =\sqrt{\left(\frac{1}{2\pi N}-0\right)^2+(0-0)^2}\\ & =\sqrt{\frac{1}{(2\pi N)^2}} \\ & =\frac{1}{2\pi N} \\ & <\varepsilon\end{align*}

This tells us that the distance from \left(\frac{1}{2\pi N},0\right) to (0,0) is less than \varepsilon, meaning \left(\frac{1}{2\pi N},0\right)\in B_\varepsilon. On the other hand, note that \sin\left(\frac{1}{\frac{1}{2\pi N}}\right)=\sin(2\pi N)=0 so that

    \[\left(\frac{1}{2\pi N},0\right)=\left(\frac{1}{2\pi N},\sin\left(\frac{1}{\frac{1}{2\pi N}}\right)\right)\]

Certainly, however, \frac{1}{2\pi N}\in(0,1], hence this implies that \left(\frac{1}{2\pi N},0\right)\in A. As \varepsilon>0 was arbitrary, this implies that B_\varepsilon\cap A\ne\varnothing regardless of choice of \varepsilon>0. As we always have that B_\varepsilon\subseteq U for some sufficiently small \varepsilon>0, this implies that U\cap A\ne\varnothing.

Now let’s return to the proof of case 2. Recall that we are assuming that B\subseteq U. Then our work has shown that A\cap V\ne\varnothing, and now that A\cap U\ne\varnothing. By following now roughly the same work as in case 1, we see that C=A\cap U, D=A\cap V form a separation of \A, contradicting the fact that A is connected

Summary: Thus in either case, we reach a contradiction. As these are the only two possible cases, we then have reached a total contradiction, and hence our assumptions must be false. Thus no such separation U,V of \Delta can exist, and \Delta must then be connected.


\Delta is not Path Connected

Theorem: \Delta is not path connected.

Proof: To show this, it will suffice to show that there is at least one pair of points which cannot be connected via a continuous path. What we do is show that no point of B can be connected to any point of A via a continuous path. However, this is a long, tedious proof, so I lay out the steps quite clearly here first, then go through the motions:

  1. Start with a path \gamma from a point in B to a point in A.
  2. Show that we can safely assume that \gamma “immediately leaves” B to go into A, instead of moving around B for a while.
  3. Show that for a small interval of x-values, all points of A are within a certain distance from a point on B.
  4. Find specific points on A within this interval that are not the necessary distance from the point on B, reaching a contradiction.

So suppose otherwise, i.e. suppose that \Delta is path connected. Then certainly for some point \mathbf{x}\in B and some \mathbf{y}\in A there is a path [0,1]\xrightarrow{\gamma}\Delta such that \gamma(0)=\mathbf{x} and \gamma(1)=\mathbf{y}. Note that we may also write \gamma=(\gamma_1,\gamma_2) where \gamma_1,\gamma_2 are the component functions of \gamma, i.e. \gamma_1=\pi_x\circ\gamma, where \pi_x is the projection of \mathbb{R}^2 onto the x-axis, i.e. \mathbb{R}^2\xrightarrow{\pi_x}\mathbb{R}, (x,y)\longmapsto x. Additionally, note that \gamma is continuous if and only if \gamma_1,\gamma_2 are continuous, so all three are continuous.

Next, our claim is that we may safely assume that \sup\left\{t\in[0,1]:\gamma\big([0,t]\big)\subseteq B\right\}=0. Intuitively, you may think about this as meaning that the path \gamma “immediately leaves” B and enters A (though it may go back to B at any point). To see this, note first that X=\left\{t\in[0,1]:\gamma\big([0,t]\big)\subseteq B\right\} is compact. Certainly it is bounded as X\subseteq [0,1]. To see that it is closed, suppose that t' is a limit point of X. By definition, this tells us that we have a sequence t_n of points in X such that \lim_{n\to\infty}t_n=t'. Moreover, we may suppose that t_n<t' for all n, as if there were some N such that t'\le t_N, then this implies [0,t']\subseteq[0,t_N] and hence that \gamma\big([0,t']\big)\subseteq\gamma\big([0,t_N]\big)\subseteq B, hence t'\in X. Note that this implies that for any t''\in[0,t') that \gamma\big([0,t'']\big)\subseteq B; for suppose there were some t''\in[0,t') such that \gamma\big([0,t'')\big)\not\subseteq B. As t_n\to t', this implies that there is some N such that t''<t_N<t' (as we are assuming t_n<t' for all n), and we know that [0,t'']\subseteq[0,t_N], which implies \gamma\big([0,t'']\big)\subseteq\gamma\big([0,t_N]\big)\subseteq B, contradicting the assumption that \gamma\big([0,t'']\big)\not\subseteq B. Thus we see that \gamma\big([0,t')\big)\subseteq B; to show that \gamma\big([0,t']\big)\subseteq B – and hence that t'\in X – it will suffice to show that \gamma(t')\in B. As each t_n\in X, this implies that \gamma\big([0,t_n]\big)\subseteq B, so that \gamma(t_n)\in B for each n\in\mathbb{N}. However, \gamma is continuous by assumption, hence we must have that \gamma(t')=\gamma\left(\lim_{n\to\infty}t_n\right)=\lim_{n\to\infty}\gamma(t_n). However, B is closed, and clearly \gamma(t') is then a limit point of B, so we must have that \gamma(t')\in B. Thus in totality we must have that \gamma\big([0,t']\big)\subseteq B and hence that t'\in X; as t' was an arbitrary limit point of X this implies that X contains all of its limit points and is thus closed. As X is then closed and bounded, it is compact.

Then let t_0=\sup X and suppose that t_0>0. As X is compact, t_0\in X, so that \gamma(t_0)\in B. Then consider [0,1]\xrightarrow{\sigma}[t_0,1] given by \sigma(t)=(1-t_0)t+t_0; composing \gamma\big|_{[t_0,1]}\circ\sigma then gives us another path in \Delta, call it \eta. However, note that \eta(0)=\gamma(t_0)\in B and \eta(1)=\gamma(1)=\mathbf{y}\in A, so \eta is also a path in \Delta starting in B and ending in A. Moreover, \sup\left\{t\in[0,1]:\eta\big([0,t]\big)\subseteq B\right\}=0. For suppose that t_1=\sup\left\{t\in[0,1]:\eta\big([0,t]\big)\subseteq B\right\}>0. By tracing back out definitions, this in turn says that \gamma\big([t_0,(1-t_0)t_1+t_0]\big)\subseteq B, so that \gamma\big([0,(1-t_0)t_1+t_0]\big)\subseteq B. However, as t_1>0 by assumption, this implies that t_0<(1-t_0)t_1+t_0, and this contradicts t_0 being the supremum of X. Thus we must have that t_1=0. As such, by replacing \gamma by \eta if t_0>0, we may safely assume that \sup\left\{t\in[0,1]:\gamma\big([0,t]\big)\right\}=0. In turn, this implies that for any 0<\epsilon<1, there is some t\in[0,\epsilon] such that \gamma(t)\in A (else 0 is not the supremum of X).

Now, before continuing, there is something we must make clear about the process. We started with an arbitrary path \gamma starting at \mathbf{x}\in B and ending at \mathbf{y}\in A. In the above work, we potentially modified \gamma so that it has the property that for any 0<\epsilon<1 there is some 0<t<\epsilon such that \gamma(t)\in A. In the process of doing so, however, we may no longer have that \gamma(0)=\mathbf{x}. However, we still have that \gamma(0)\in B, meaning we still have a path starting at a point in B and ending at a point in A. So by relabeling, we may still assume that \gamma(0)=\mathbf{x}\in B and \gamma(1)=\mathbf{y}\in A, even if these are not the same points we started by choosing. To sum up; we started by assuming the existence of a certain path, and used this to build a new path with desirable properties; now we show that this new path causes problems.

Again, let \gamma=(\gamma_1,\gamma_2), and let \gamma(0)=\mathbf{x}=(0,y_0) (note that the x-coordinate may safely be assumed to be 0 as \mathbf{x}\in B). As \gamma_2 is continuous, there is some \delta>0 so that for all t\in[0,1] with |t-0|<\delta, |\gamma_2(t)-y_0|<\frac{1}{2}. Note that this is the same as saying for all t\in[0,1] with t<\delta, |\gamma_2(t)-y_0|<\frac{1}{2}. Then pick some t'\in[0,1] with 0<t'<\delta such that \gamma(t')>0; note that we can do this as if we take \epsilon=\delta, then there is some t'\in(0,\epsilon) such that \gamma_1(t')\in A, and the x-coordinate of any point in A (in this case, \gamma_1(t')) is always positive. Now consider the interval [0,t']; applying \gamma_1 gives us \gamma_1\big([0,t']\big), and this must be connected as [0,t'] is connected as an interval, and \gamma_1 is continuous. However, \gamma_1\big([0,t']\big)\subseteq\mathbb{R}, and as shown before any connected subset of \mathbb{R} must be an interval. Therefore, \gamma_1\big([0,t']\big) is an interval. Moreover, \gamma_1(0)=0 and \gamma_1(t')>0, so \gamma_1\big([0,t']\big) must contain the interval [0,\gamma_1(t')].

Now let’s think about what this means. Take any x\in(0,\gamma_1(t')]; as (0,\gamma_1(t')]\subseteq \gamma_1\big([0,t']\big), this implies that there is some t\in[0,t'] such that \gamma_1(t)=x; as x>0 this implies that \gamma(t)\in A so that in turn we must have that \gamma(t)=\left(x,\sin\left(\frac{1}{x}\right)\right). However, as t\le t'<\delta, this implies that |\gamma_2(t)-y_0|<\frac{1}{2}, but \gamma_2(t)=\sin\left(\frac{1}{x}\right), so that \left|\sin\left(\frac{1}{x}\right)-y_0\right|<\frac{1}{2}. Essentially, this means that for some small interval on the x-axis, there is a point on A who’s y-value is at most \frac{1}{2} a unit away from y_0. But of course, as x gets very close to 0, \sin\left(\frac{1}{x}\right) oscillates quite quickly, so that there will always be a point on A that violates this.

Finally, the specific contradiction. To do this, there are two cases to consider. Case 1: y_0\ge 0. Take any N\in\mathbb{N} such that 2\pi N-\frac{\pi}{2}>\frac{1}{\gamma_1(t')}. In turn, this implies that \frac{1}{2\pi N-\frac{\pi}{2}}<\gamma_1(t') so that \frac{1}{2\pi N-\frac{\pi}{2}}\in(0,\gamma_1(t')]. By the preceding paragraph, this implies that

    \[\left|y_0-\sin\left(\frac{1}{\frac{1}{2\pi N-\frac{\pi}{2}}}\right)\right|<\frac{1}{2}\]


    \begin{align*}\left|y_0-\sin\left(\frac{1}{\frac{1}{2\pi N-\frac{\pi}{2}}}\right)\right|& = \left|y_0-\sin\left(2\pi N-\frac{\pi}{2}\right)\right| \\ & = \left|y_0-\sin\left(-\frac{\pi}{2}\right)\right| \\ & =\left|y_0-(-1)\right| \\ & = \left|y_0+1\right| \\ & = y_0+1 \\ & >1\end{align*}

a contradiction. Case 2: y_0<0. Then take any N\in\mathbb{N} such that 2\pi N+\frac{\pi}{2}>\frac{\gamma_1(t')}. In turn, this implies that \frac{1}{2\pi N+\frac{\pi}{2}}<\gamma_1(t') so that \frac{1}{2\pi N+\frac{\pi}{2}}\in(0,\gamma_1(t')]. Again, we then must have that

    \[\left|y_0-\sin\left(\frac{1}{\frac{1}{2\pi N+\frac{\pi}{2}}}\right)\right|<\frac{1}{2}\]


    \begin{align*}\left|y_0-\sin\left(\frac{1}{\frac{1}{2\pi  N+\frac{\pi}{2}}}\right)\right|& = \left|\sin\left(\frac{1}{\frac{1}{2\pi  N+\frac{\pi}{2}}}\right)-y_0\right|\\ &  \left|\sin\left(2\pi  N+\frac{\pi}{2}\right)-y_0\right| \\ & =  \left|\sin\left(\frac{\pi}{2}\right)-y_0\right| \\ &  =\left|1-y_0\right| \\ & = 1-y_0 \\  & >1\end{align*}

a contradiction. Thus in both possible cases we’ve reached a contradiction. As such, not such path \gamma can exist, and so \Delta is not path connected.


Topology Explained – Seeing Double: The Line with Two Origins

Sometimes, when I have nothing to do late weekend nights, I have a few drinks and watch Bob Ross in The Joy of Painting. I don’t know how to paint, but it’s soothing and helps me fall asleep. Anyways, sometimes I have a few too many drinks, and all of a sudden there’s two of everything: two trees, two clouds, two mountains, two Bobs (if only). Anyone who’s knocked a few too many back is probably familiar with the idea. One problem with the human eye – at least, while heavily intoxicated – is that everything goes out of focus; it’s not just the periphery of your vision, or the very center, that goes double, it’s your entire field of view. But imagine if it weren’t like this; imagine if at an infinitesimally small point directly in the center of your vision, you saw double, but everything else was crystal clear. Now imagine you stare at a line while this is happening; you’d likely see something like this:

If you’re actually seeing something like this, don’t drive anywhere.

This is what is known as The Line with Two Origins (henceforth simply the line), and I hate it. And not just because I see things like it after too many drinks; I hate it because it is almost always drawn wrong. Before I complain too much, here are two other common ways I see the line with two origins drawn:

Slightly better, but still bad.
Again, slightly better than the last, but still not quite right.

Now let me explain my issues with these drawings, and then we’ll walk through building the damn thing and explaining why it’s interesting. Fundamentally my issues with these drawings come from the following two misconceptions:

  1. The “axis” appears “broken”, i.e. in the first two drawings there’s a gap in the “axis”. In reality, the “axis” is not only kept together in the line, but the line itself is path connected, meaning there’s a continuous way to get from any point on the line to any other point, including the two “origins”.
  2. The two “origins” appear to be disparate points, or otherwise some distance apart. In reality, however, the two “origins” are infinitely close to each other; in some sense even though they are “different” points, they occupy exactly the same “place” on the “axis” of the line.

At the end of this, I’ll propose a drawing that I think makes the most sense to me, but I make no guarantees about its accuracy. In any case, let’s dig a bit more into what makes the line special, then go about building it.

What’s So Special About The Line Anyways?

Other than my hatred for how it’s typically drawn, there are two things about the line that make it “special” relatively speaking. First, it’s perhaps one of the “nicest” topological spaces which is not Hausdorff. For those a bit more well-versed in differential geometry terms, it is also the prototypical example of a manifold which is not Hausdorff. Second, and more pertinent to my own interests, is that the line is not only connected, but path connected. I find this interesting, as you would never guess this just looking at one of the “standard” drawings of the line. Now that we know what’s special about the line, let’s get to building it.

Setup and Construction

Denote by \mathbb{R}_1 the set


Similarly, denote by


So that \mathbb{R}_1 and \mathbb{R}_{-1} are the horizontal lines at y=1,-1, respectively. Likewise, let \mathcal{L}=\mathbb{R}_1\cup\mathbb{R}_{-1}, so that \mathcal{L} is simply these two lines viewed as a subspace of \mathbb{R}^2.

To proceed with the construction of the line with two origins, you as the reader need to be familiar with the notion of quotients, quotient maps, and the quotient topology. You can find all of the requisite information in my post on making circles out of lines.

In any case, define a relation \sim on \mathcal{L} by saying that

    \[(x,y)\sim(r,s)\)\Longleftrightarrow\begin{array}{c} (x,y)=(r,s) \\ \text{ or } \\ x=r\text{ and }x\ne0\end{array}\]

As I always do when I define an equivalence relation, I leave it you to verify that it is indeed an equivalence relation. What this equivalence relation is then doing is identifying points on \mathbb{R}_1 and \mathbb{R}_{-1} with identical x-values except those on the y-axis:

Let \mathcal{L}\xrightarrow{\pi}\mathcal{L}/\sim be the usual quotient map with \mathcal{L}/\sim given the quotient topology; for short, call \mathcal{L}/\sim=L. But now let’s try and picture what L should “look like” (technically speaking, there’s no reason it should “look like” anything, but oh well, topology must go on): when quotienting out by the relation \sim, we are effectively smashing all related points together in a “continuous” way. Here’s one way of thinking about the process pictorially, though this does not accurately reflect what’s actually going on (without proof, anyways):

Getting close ..

It should hopefully then be no surprise what we end up with:


Yep, L is actually just the line. Moreover, it’s easy to see how we get the above picture of the line when we think about the construction of the line as we have.

Before we go any further, let’s adopt some notation: let p=[(0,1)] and q=[(0,-1)]; as you might imagine, these are the two “origins” of the line, and will play an important role. Moreover, let \Lambda=L\setminus\{p\} and \Omega=L\setminus\{q\}; pictorially you may think about these as:

I am the Lambda …
and the Omega

There’s something crucial that the above pictures really miss: the topology of the line, and this is what we’re going to dig into. To do this, let’s first look a little bit more closely at the map \pi.

The Quotient Map \pi

We know that \pi is a quotient map, simply by construction. But what else can be said about it? There’s one key fact about \pi that will be useful to prove, but first, let’s take a look at what open sets in L “look like”. Recall that a subset V\subseteq L is open if and only if \pi^{-1}(V) is open in \mathcal{L}. Imagine we take an “interval” U on L away from the two origins:

Now “lifting” this back up to \mathcal{L} via \pi we see that it “lifts” to two “intervals” on \mathbb{R}_{1},\mathbb{R}_{-1}:

Going Up

Note, however, that these intervals are actually open subsets of \mathcal{L}, since we can view them as open balls in \mathbb{R}^2 intersected with \mathcal{L}:

Now this shouldn’t come as a shock; should U be open it should be pulled back to an open subset of \mathcal{L} as \pi is continuous. However, we can also think about going the other direction; suppose we start with an “interval” W on \mathcal{L}:

W for Wumbo

Now if we “push W down” to the quotient L, we get:

Going Down

But, at least in this picture argument, we already know that \pi(W) is open in L as it “lifts” to open “intervals” in \mathcal{L}. Thus away from the origins p,q, it appears as though \pi is actually an open map, i.e. it sends open sets to open sets. However, the unfortunate fact about the line is that ll of the “interesting” behavior occurs around the origins. So suppose we take an open “interval” about the point (0,1) in \mathcal{L}:

Here comes trouble

When we “push” this down via \pi, we again get an “interval” in L:

Going Down again

The problem comes when we pull this back to \mathcal{L} via \pi; we of course get back our original “interval”, but we also get back a “broken” interval on \mathbb{R}_{-1}:

Take the elevator up

However, a moments thought says that this is still OK: we can get this punctured interval by intersection \mathbb{R}_{-1} with a punctured open disk centered at (0,-1) (at least for sufficiently small intervals; for bigger ones just use a punctured rectangle or something). Thus it would appear that \pi indeed is an open map. In fact, this is true, \pi is indeed an open map. However, what we have talked about certainly does not constitute a proof of this fact, but it does cover the “big ideas” behind the proof. However, we only need the fact that \pi is an open map for one small argument, and even then I view this argument as a minimally important one; moreover, the actual proof that \pi is an open map is quite tedious, requiring considering several cases to be thorough. Thus at this time I am choosing to omit the proof that \pi is an open map, and instead use it freely a little bit later (sorry).

What is more important about our discussion above is that it gives a good idea for why the origins p,q are “infinitely close” to each other. For suppose I take any neighborhood of p in L; for simplicity’s sake we may as well assume that it is some “interval” in L. Now any such interval of p will lift to an interval of (0,1) in \mathcal{L}, but it will also lift to some “broken” interval about (0,-1) in \mathbb{R}_{-1} as we saw above. But any neighborhood of q in L will also contain some small piece of this “broken” interval in \mathbb{R}_{-1} once lifted by \pi. Applying \pi again, we see that any neighborhood of p and any neighborhood of q must have some overlap in L so that p and q are “inseparable” by open neighborhoods. In other words, L is not Hausdorff. We will prove this more rigorously in a moment. First however, let’s prove what L is.

\Lambda,\Omega are homeomorphic to \mathbb{R}

Let’s go back and talk about \Lambda=L\setminus\{p\} and \Omega=L\setminus\{q\}. These will play big roles in the results that follow. The key fact about these subspaces are that they are, in fact, open, and that each is homeomorphic to the real line \mathbb{R}.

Proposition: \Lambda,\Omega are open subspaces of L.

Proof: We only show that \Lambda is open, as the argument for \Omega is analogous. Then to show that \Lambda is open in L, it will suffice to prove that \pi^{-1}(\Lambda) is open in \mathcal{L}. The claim is that


Then take any (x,y)\in\pi^{-1}(\Lambda); this implies that \pi(x,y)\in\Lambda. There are two possibilities. If x\ne0, then certainly (x,y)\ne(0,1), so (x,y)\in\mathcal{L}\setminus\{(0,1)\}. Else if x=0 and \pi(x,y)\in\Lambda, we must have that (x,y)=(0,-1), and again we get that (x,y)\in\mathcal{L}\setminus\{(0,1)\}.

Conversely, take any (x,y)\in\mathcal{L}\setminus\{(0,-1)\}. There are then two options: x=0 or x\ne0. If x\ne0, then certainly (x,y)\not\sim(0,1), so \pi(x,y)\ne[(0,1)], and hence \pi(x,y)\in\Lambda. Thus x\in\pi^{-1}(\Lambda). Thus the two sets are equal, as claimed.

Finally, simply note that \mathcal{L}\setminus\{(0,1)\} is open in \mathcal{L} as, for instance,


and certainly \mathbb{R}^2\setminus\{(0,1)\} is open in \mathbb{R}^2.


Now for the meat of the section:

Theorem: \Lambda\cong\mathbb{R}\cong\Omega

Proof: Again, we show only that \Lambda\cong\(\mathbb{R}, as the argument for \Omega is almost identical. Define a map L\xrightarrow{\psi}\mathbb{R} by

    \[\psi\big([(x,y)]\big) = x\]

Note that this is well defined, as if we are given [(x,y)]\in L, either x\ne0 and hence [(x,y)]=\{(x,y),(x,-y)\} and any potential representative has the same x-value, or [(0,1)]=\{(0,1)\},[(0,-1)]=\{(0,-1)\} and there is no ambiguity with our choice of representative. Next, note that \psi is certainly continuous. To show this, we use that fact that this map is continuous if and only if its composition with the quotient map \pi is continuous, as L is, after all, a quotient space. The claim is then that for any (x,y)\in\mathcal{L} we have that


But again, this follows immediately from the definitions of \pi,\psi. But this is certainly continuous, as this is simply the restriction of the projection \mathbb{R}^2\to\mathbb{R} given by (x,y)\longmapsto x to \mathcal{L}. Thus since the composite \psi\circ\pi is continuous, \psi is also continuous. Restricting \psi to \Lambda then gives us a continuous map \Lambda\to\mathbb{R}. It is relatively straightforward to check that this is indeed bijective when restricted to \Lambda, with inverse \mathbb{R}\xrightarrow{\psi^{-1}}\Lambda given by \psi^{-1}(x)=[(x,-1)]. As \psi was already proven continuous, we need only prove that \psi^{-1} is continuous to prove that \psi is a homeomorphism, and finish the proof. This, however, is also fairly immediate. You can think of \psi^{-1} as the following composition: \mathbb{R}\xrightarrow{\iota}\mathcal{L}\xrightarrow{\pi}\Lambda where \iota is the restriction of the continuous inclusion map \mathbb{R}\to\mathbb{R}^2 given by x\longmapsto (x,-1) to \mathcal{L}, and \pi is simply the restriction of the quotient map to \Lambda. Of course you should verify that this is indeed the case, and that these restrictions are well founded. As both these restrictions of \pi,\iota are themselves continuous, this implies that \psi^{-1} is also continuous. Thus \psi is a homeomorphism, and form this we get that \Lambda\cong\mathbb{R}.


L is (almost) a Manifold

Most mathematicians agree that a sensible definition for a manifold is the following: given a space M, M is a manifold if

  1. M is locally euclidean, i.e. if for each x\in M there is a neighborhood U_x of x such that U_x is homeomorphic to \mathbb{R}^n for some n.
  2. M is second countable, i.e. M has a countable basis.
  3. M is Hausdorff.

I’ve listed the manifold criterion in roughly order of importance. The “big idea” behind a manifold is that it locally “looks like” euclidean space. Second countability allows one to use useful gadgets like partitions of unity. Hasdorffness just keeps you sane. Let’s see which of these L manages to tick off.

Proposition: L is locally euclidean.

Proof: First, note that \Lambda\cup\Omega=L. Take some x\in L; either x\in \Lambda or x\in \Omega. However, in either case, we know that \Lambda or \Omega will be a neighborhood of x as we’ve already shown they’re open, and \Lambda\cong \mathbb{R}\cong\Omega so that both are homeomorphic to \mathbb{R}. As x\in L was arbitrary, this finishes the proof.


Proposition: L is second countable.

I have decide not to prove this at the current time, as it does not, I believe, really get to the spirit of L being (almost) a manifold. However, it is true, and I outline the idea here. The basic idea is that if X is a second countable space and Y is a subspace of X, then Y is also second countable by simply intersecting the countable base with Y. Now \mathbb{R}^2 is certainly second countable, as for instance you could take as a base all open balls with rational centers and rational radii. As \mathcal{L} is a subspace of \mathbb{R}^2 it too is then second countable. Now it is not in general true that a quotient of a second countable space need be second countable, however if X\xrightarrow{\pi}Y is an open quotient map, then Y is a second countable quotient of X, simply by pushing forward the countable base of X via \pi. As we have “proven” previously that \mathcal{L}\xrightarrow{\pi}L is an open quotient map, this in turn implies that L is also second countable.

Well, we’ve shown that L satisfies the first two requirements for being a manifold, but we claim that it is not a manifold. This must then mean that …

L is not Hausdorff

Let’s just get right into it.

Theorem: L is not Hausdorff.

Proof: The problem should obviously come at the origins, p,q. Then take any neighborhood U of p and V of q in L. As \pi is continuous, \pi^{-1}(U) and \pi^{-1}(V) will be open in \mathcal{L}. However, (0,1)\in\pi^{-1}(U) and (0,-1)\in\pi^{-1}(V) as \pi(0,1)=[(0,1)]=p\in U and likewise \pi(0,-1)=[(0,-1)]=q\in V. However, if (0,1)\in\pi^{-1}(U) and \pi^{-1}(U) is open in \mathcal{L}, there must be some \varepsilon>0 so that \{(x,1):-\varepsilon<x<\varepsilon\}\subseteq \pi^{-1}(U), as \pi^{-1}(U) must contain the intersection of a sufficiently small ball about (0,1) with \mathcal{L}. An analogous argument shows that there is some \varepsilon'>0 such that \{(x,-1):-\varepsilon'<x<\varepsilon'\}\subseteq\pi^{-1}(V); by shrinking one of \varepsilon,\varepsilon' if necessary, we may assume that these are the same. Thus we see that, for instance, \left(\frac{\varepsilon}{2},1\right)\in\pi^{-1}(U) and likewise that \left(\frac{\varepsilon}{2},-1\right)\in\pi^{-1}(V). In turn, this implies that \left[\left(\frac{\varepsilon}{2},1\right)\right]=\pi\left(\frac{\varepsilon}{2},1\right)\in U and \left[\left(\frac{\varepsilon}{2},-1\right)\right]=\pi\left(\frac{\varepsilon}{2},-1\right)\in V. However, as \frac{\varepsilon}{2}>0, we know that \left(\frac{\varepsilon}{2},1\right)\sim\left(\frac{\varepsilon}{2},-1\right), so that \left[\left(\frac{\varepsilon}{2},1\right)\right]=\left[\left(\frac{\varepsilon}{2},-1\right)\right]. In turn, this implies that \left[\left(\frac{\varepsilon}{2},1\right)\right]\in U\cap V, so that U\cap V\ne\varnothing. But U was an arbitrary neighborhood of p and V was an arbitrary neighborhood of q, and so this proves that any neighborhood of p and any neighborhood of q will have non-trivial intersection. Thus, L is not Hausdorff.


L is Path Connected

This is ultimately the second of my problems with the usual drawings of the line; very often it looks like there’s no way the space is path connected, yet L is almost indistinguishable from \mathbb{R} itself. To prove this, we need the following little lemma:

Lemma: Let X be a space, and let U,V\subseteq X be open subspaces so that U\cup V=X, both U,V are path connected, and U\cap V\ne\varnothing. Then X is also path connected.

Proof: Take any x,y\in X; then either x,y\in U, x,y\in V, x\in U and y\in V, or x\in V and y\in U as U\cup V=X. In the first two cases, there’s nothing to show as U,V are both already path connected. Assume then that x\in U and y\in V; for the reversed case you can simply reverse the path we are about to construct. As U\cap V\ne\varnothing, say that z\in U\cap V. Then as U is path connected and x,z\in U there is a path \gamma connecting x to z, i.e. there is a continuous map [0,1]\xrightarrow{\gamma}U such that \gamma(0)=x and \gamma(1)=z. As V is path connected and y,z\in V this implies there is a path \eta connecting z to y, i.e. there is a continuous map [0,1]\xrightarrow{\eta}V such that \eta(0)=z and \eta(1)=y. Note that as U,V are open subsets of X we can certainly extend \gamma,\eta to continuous maps [0,1]\xrightarrow{\gamma,\eta}X with the aforementioned properties. Then define a function [0,1]\xrightarrow{\xi}X by

    \[\xi(t)=\left\{\begin{array}{cl} \gamma(2t) & : 0\le t\le\frac{1}{2} \\ \eta(2t-1) & : \frac{1}{2}\le t\le 1\right.\]

Now clearly \xi(0)=x and \xi(1)=y. Moreover \gamma(1)=z=\eta(0) so that these functions agree on their overlap at t=\frac{1}{2}. Thus if \xi is continuous, it is then a path from x to y in X. I leave the verification of continuity via various stupid gluing lemmas to you. As such, given generic x,y\in X, we were able to connect them via a path, and so X is path connected.


Now let’s see how to apply this to L. Let U=\Lambda and V=\Omega. Then certainly \Lambda\cup\Omega=L and \Lambda\cap \Omega\ne\varnothing. To apply the above lemma, we need only show that \Lambda,\Omega are path connected. However, we have already shown that \Lambda\cong\mathbb{R}\cong\Omega. Path connectedness of a space is preserved by homeomorphism, and certainly \mathbb{R} is path connected (as it’s really just a big path anyways). Thus the lemma then tells us that L is indeed path connected.


So there you have it; the line with two origins. Hopefully after working through all of these details, you share in my frustration at the deficiency of most pictures of the line with two origins. Despite being hard to visualize, the line with two origins is actually quite a nice space, especially given that it fails to be Hausdorff. It is also a common example or counter example, so having a good working understanding of it is a nifty trick to keep in your back packet in the bad parts of the mathematical world. Stay tuned for next week when we talk about the line with three origins.

Topology Explained – Tying the Knot: Making Circles from Lines

Imagine, if you will, that you have a piece of string. Here’s a visualization


Now imagine that you can “stick” the two ends of the string together, say using some chewed gum or something or other; you’ll end up with something like the following:

Trident Original Flavor; surprisingly good and underrated.

Now with a little bit of straightening out and cleaning up, we get a circle:

Yep, that’s a circle alright. Trust me, I’m a mathematician.

And there you have it; we’ve just completed one of the myriad ways that mathematicians build circles. So what’s the big hullabaloo about then? Well, there’s always a catch. What we just did is what is often referred to as intuitively obvious, and this is a very dangerous phrase indeed. Of course if you take a piece of string and tie the ends together you end up with a circle; this is something that you can very easily test yourself and something that I personally end up doing accidentally every morning while flossing. But we never proved this. Moreover, while it is universally accepted that this works out mathematically, the construction of the circle from a line is fairly obfuscating, and it is not immediately apparent that the space we build is actually the same as a circle. And indeed, the proof that the space one builds is actually the same as the circle is deceptively intricate and lengthy, even though most topologists dismiss it as being intuitively obvious. So let’s take a while and walk through the proof itself, and show that this intuitively obvious fact is anything but.


If you’re already familiar with quotient spaces and maps, feel free to skip this section.

Definition: Let X be a topological space, Y be a set, and X\xrightarrow{f}Y a surjective set map. The quotient topology on Y induced by f (or simply quotient topology if context makes this clear) is the finest topology on Y such that f is continuous.

In essence, the quotient topology is the “smallest” topology on a set Y which makes a given map continuous. This is a good intuitive description, but let’s give something a little bit more concrete.

Proposition: Let X be a space, Y a set, and X\xrightarrow{f}Y a surjective set map. The quotient topology on Y induced by f can be characterized in the following way: a subset V\subseteq Y is open in Y if and only if f^{-1}(V) is open in X.

Proof: Take any open V\subseteq Y; by virtue of the fact that the quotient topology makes f continuous, this implies that f^{-1}(V) is open in X. Conversely, suppose that the quotient topology on Y is given by \mathcal{T}. Moreover, suppose for some V\subseteq Y we have that f^{-1}(V) is open in X, but V is not open in Y. Then let \mathcal{T}' denote the collection of subsets of Y that you can obtain via unioning any collection of sets in \mathcal{T} with V or amongst themselves, and by intersecting any finite collection of sets of \mathcal{T} with V or amongst themselves. Then the claim is that \mathcal{T}' is also a topology on Y. I’ll leave the excruciating verification of this to you. However, note that certainly \mathcal{T}\subseteq\mathcal{T}' as any element of \mathcal{T} can be written as the union with itself. Thus \mathcal{T}' is a finer topology on Y than the quotient topology; if we can show that f is still continuous when equipping Y with \mathcal{T}' then we will have reached a contradiction as we are assuming that the quotient topology is the finest topology on Y with this property. Then take any U\in\mathcal{T}'; if U\in\mathcal{T} then there’s nothing to show about f^{-1}(U) as we already know it’s open by definition of the quotient topology. Thus we really need only consider two cases: when U=\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V where U_\alpha\in\mathcal{T} for each \alpha\in I, and U=\left(\bigcap_{i=1}^nU_i\right)\cap V where U_i\in\mathcal{T} for each i, as these are the only other possibilities for elements of \mathcal{T}'

Then suppose that U=\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V where U_\alpha\in\mathcal{T} for each \alpha\in I. By definition, we then get that

    \[f^{-1}\left(U\right)=f^{-1}\left(\bigcup_{\alpha\in I}U_\alpha\right)\cup V\right)=\left(\bigcup_{\alpha\in I}f^{-1}(U_\alpha)\right)\cup f^{-1}(V)\]

By our assumptions, however, each f^{-1}(U_\alpha) is open in X by definition of the quotient topology, and f^{-1}(V) is open in X by our assumption on V. Thus f^{-1}(U) is open in X as the union of open sets.

Otherwise, suppose that U=\left(\bigcap_{i=1}^nU_i\right)\cap V where U_i\in\mathcal{T} for each i. Then

    \[f^{-1}(U)=f^{-1}\left(U=\left(\bigcap_{i=1}^nU_i\right)\cap V\right)=\left(\bigcap_{i=1}^nf^{-1}(U_i)\right)\cap f^{-1}(V)\]

But again, by similar reasoning each f^{-1}(U_i) and f^{-1}(V) are open, so f^{-1}(U) is then open as the intersection of finitely many open sets.

Thus, to recap our argument, we’ve shown that for any U\in\mathcal{T}', f^{-1}(U) is open in X, hence f is continuous when Y is equipped with the topology {\mathcal{T}'. However, \mathcal{T}' is finer than that quotient topology, \mathcal{T}, which is the finest topology on Y such that f is continuous, a contradiction. Thus no such V\subseteq Y where f^{-1}(V) is open in X but V is not open in Y can exist. As such, if f^{-1}(V) is open in X, we must have that V is open in Y, completing the proof.


Let’s talk about the most common way quotient topologies come up. Suppose that we have a space X, and an equivalence relation \sim on X. Let X/\sim denote the set of the equivalence classes of \sim, and let X\xrightarrow{\pi}X/\sim be the “canonical” surjection sending an element x to its equivalence class [x], i.e. \pi(x)=[x]. Then we can equip X/\sim with the quotient topology induced by \pi; it turns out that this is most often the “natural” topology to put on the space X/\sim. This comes up most often in two contexts: “gluing” points together, and “collapsing” regions. Creating the circle from an interval is an example of the former; you “glue” the ends of the interval together to form the circle. For the latter, you often take a connected region in a space X, say A, and collapse it to a single point to form the quotient X/A, where you can think about this notation as saying we are “modding out” the region A and “reducing” it to a single point. I may eventually do a post or two about this construction, likely in the context of the ambiguity behind the notation \mathbb{R}/\mathbb{Z}.

In any event, there are a few more results we should talk about with regards to quotients. 

Definition: Let X,Y be spaces. Given a surjective continuous map X\xrightarrow{f}Y, we say that f is a quotient map if for any V\subseteq Y, f^{-1}(V) being open in X implies that V is open in Y. In light of the preceding proposition, we see that f is then a quotient map if and only if Y has the quotient topology induced by f.

Proposition: Let X\xrightarrow{f}Y be a quotient map. Let Z be a space, and let Y\xrightarrow{g}Z be a set map, so that we are in the following situation:

Mmm, Commutative Diagrams

Then the map g is continuous if and only if the composition g\circ f is continuous.

Proof: Suppose first that g is continuous. Then g\circ f is also continuous, as the composition of continuous maps. Conversely, suppose that g\circ f is continuous. Take any W\subseteq Z open; as g\circ f is continuous, this implies that \big(g\circ f\big)^{-1}(W) is open in X. However,

    \[\big(g\circ f\big)^{-1}(W)=f^{-1}\big(g^{-1}(W)\big)\]

As such, by our work on the quotient topology, if f^{-1}\big(g^{-1}(W)\big) is open in X, we know that g^{-1}(W) is open in Y. As W\subseteq Z was arbitrary, this implies that g is continuous.


Setup and Construction

Let’s take a moment to establish some notation and facts. First, we’ll set the unit interval to be I=[0,1]. Second, the unit circle will be \mathbb{S}^1=\left\{(x,y)\in\mathbb{R}^2:\sqrt{x^2+y^2}=1\right\}.

Next we want to go through the process of “building” the circle via identifying the endpoints of I. To do this, we make use of the quotient topology. Define a relation \sim on I by

    \[x\sim y\Longleftrightarrow \begin{array}{c} x=y \\ \text{or} \\ x=0 \text{ and }y=1 \\ \text{or} \\ x=1\text{ and }y=0\end{array}\]

I will leave it to you to verify that this is an equivalence relation on I with the following properties: for any x\in I with 0<x<1, x is only related to itself, and the only other relation is that 0\sim 1. Thus for any x\in I with 0<x<1 we have that [x]=\{x\}, and [0]=\{0,1\}=[1].

Now let C=I/\sim, and equip C with the quotient topology induced by the “canonical” surjection I\xrightarrow{\pi}C. The claim is that we are now done; the space C is simply the circle, \mathbb{S}^1. Now hopefully you see my issue with topologists; we have constructed this vague space C and are simply claiming that, yes, indeed C is indistinguishable from the unit circle \mathbb{S}^1 itself. As an undergraduate, this bothered me quite a bit as the elements of C are not even ordered pairs, so how could this be \mathbb{S}^1? Too often professors would reply that, “ah of course C is not \mathbb{S}^1 on the nose, it is simply homeomorphic to \mathbb{S}^1.” When pressed to show that these two spaces are homeomorphic, most just fell back on the argument that, hey, it’s obvious that they’re the same space, so clearly they’re homeomorphic. Hopefully now you’re starting to see some of my frustration. In any case, let’s show that C is homeomorphic to \mathbb{S}^1.


This one is a real doozy, and part of the reason I hate that people generally state that this result is “obvious”. The proof of this requires multiple parts, which we prove at first, and then bring it all together in one final proof. The fundamental result we need is the following theorem:

Theorem: Let X,Y be spaces and X\xrightarrow{f}Y a continuous bijection. If X is compact, and Y is Hausdorff, then f is a homeomorphism.

Proof: It suffices to prove that f is an open map; an open bijection has a continuous inverse, hence as f is already continuous this will imply that f is a homeomorphism. Then take any U\subseteq X open. As U is open, U^C is closed; however every closed subspace of a compact space is also compact, so as X is compact, U^C is compact. The image of a compact space under a continuous map is compact, hence f\big(U^C\big) is a compact subspace of Y, as we assume that f is continuous. But Y is Hausdorff, and every compact subspace of a Hausdorff space is closed, thus f\big(U^C\big) is a closed subspace of Y. Note that f\big(U^C\big)=\big(f(U)\big)^C, so \big(f(U)\big)^C is closed in Y. By definition, this implies that f(U) is open in Y. As U\subseteq X was arbitrary, this implies that f is an open map and hence that f is a homeomorphism.


With this in hand, our overall structure/goal is obvious: show that C is compact, \mathbb{S}^1 is Hausdorff, and then build a continuous bijection C\xrightarrow{f}\mathbb{S}^1.

Lemma: \mathbb{S}^1 is Hausdorff.

Proof: This follows immediately as any subspace of a Hausdorff space is also Hausdorff (prove this yourself if you find yourself asking why), \mathbb{S}^1\subseteq\mathbb{R}^2, and \mathbb{R}^2 is Hausdorff (as you can always take sufficiently small open balls about two distinct points).


Lemma: C is compact.

Proof: Recall our quotient map: I\xrightarrow{\pi}C; this is a continuous surjection, so that \pi(I)=C. However, \pi is continuous, and I=[0,1] is compact, so \pi(I)=C is also compact as the image of a compact space under a continuous map.


With this all done, we simply need to build a continuous bijection C\xrightarrow{f}\mathbb{S}^1. We construct the map here, then prove it has the desired properties. First, define a map C\xrightarrow{\phi}[0,2\pi) by

    \[\phi\big([x]\big)=\left\{\begin{array}{cl} 2\pi x & : 0<x<1 \\ 0 & :x=0,1\end{array}\right.\]

Note that this is well-defined, for if 0<x<1 then [x]=\{x\} so there’s only one choice of representative, and likewise \phi\big([0]\big)=0=\phi\big([1]\big). Notice that this is also a bijection; it is clearly surjective by construction, for if we take any y\in[0,2\pi, then \left[\frac{y}{2\pi}\right] maps to y under \phi. Moreover, it is injective. For suppose that [x],[y]\in C are such that \phi\big([x]\big)=\phi\big([y]\big). There are two cases to consider: \phi\big([x]\big)=0 and \phi\big([x]\big)\ne0. Suppose the latter first; this then implies that 0<x,y<1 and 2\pi x=2\pi y, so that x=y and hence [x]=[y]. In the former, this implies that x=0,1 and y=0,1; but regardless of the possible combinations we see that this implies x\sim y so that [x]=[y]. Thus in either case we see that if \phi\big([x]\big)=\phi\big([y]\big), then [x]=[y] so that \phi is injective, and hence bijective.

Next, define a map [0,2\pi)\xrightarrow{\psi}\mathbb{S}^1 by \psi(\theta)=\big(\cos(\theta),\sin(\theta)\big). Then this is again a bijection, the verification of which I leave to you (it’s really just algebra involving \sin‘s and \cos‘s, but it’s annoying algebra so you do it).

Composing \psi and \phi, we thus get a bijection C\xrightarrow{\psi\circ\phi}\mathbb{S}^1, call it \gamma=\psi\circ\phi. To finish the proof, it will suffice to prove that \gamma is continuous. To do this, recall from our work on quotients that it will suffice to prove that the composition I\xrightarrow{\gamma\circ\pi}\mathbb{S}^1 is continuous. However, the claim is that the composition \gamma\circ\pi is given by

    \[\big(\gamma\circ\pi\big)(x)=\big(\cos(2\pi x),\sin(2\pi x)\big)\]

For x\in I with 0\le x<1, this is immediate if you track down all of our definitions. However, \sin,\cos are 2\pi periodic so even though we formally have that


we in turn get that

    \[\big(\gamma\circ\pi\big)(1)=\big(\cos(0),\sin(0)\big)=\big(\cos(2\pi\cdot 1),\sin(2\pi\cdot 1)\big)\]

as needed. However, \gamma\circ\pi is then continuous, as is is simply the restriction of the map I\to\mathbb{R}^2 given by x\longmapsto\big(\cos(2\pi x),\sin(2\pi x)\big) to \mathbb{S}^1, which is continuous as its component functions, \cos(2\pi x),\sin(2\pi x) are both continuous.

Thus as \gamma\circ\pi is continuous, we then know that \gamma is continuous; as such we have found a continuous bijection C\xrightarrow{\gamma}\mathbb{S}^1. As C is compact and \mathbb{S}^1 is Hausdorff, this implies by the first theorem that \gamma is in fact a homeomorphism, and hence that C\cong\mathbb{S}^1.


And there you have it folks, a circle. This process underlines my fundamental displeasure with topology; so much of what is deemed an “intuitive” proof is omitted, and further is anything but “intuitive”. At the same time, arguing like this is necessary in topology; should every topologist be asked to argue every result as I did above, we would probably be decades, if not centuries, behind where we are now. Just think about it; it took us probably four or five pages worth of work, presupposing a good basic knowledge of point-set topology, to simply argue that tying the ends of a string together gives you a circle. Think about how much work it would take then to prove something like the Hairy Ball Theorem. And this is why hyper-specialization in math is not only good, but necessary; there’s no way in Hell I would ever research topology, so let’s be glad it’s some other poor bastards’ jobs to do it.