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.


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