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

    \[\mathbb{R}_1=\left\{(x,1)\in\mathbb{R}^2:x\in\mathbb{R}\right\}\]

Similarly, denote by

    \[\mathbb{R}_{-1}=\left\{(x,-1)\in\mathbb{R}^2:x\in\mathbb{R}\right\}\]


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:

Surprise!

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

    \[\pi^{-1}(\Lambda)=\mathcal{L}\setminus\{(0,1)\}\]

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,

    \[\mathcal{L}\setminus\{(0,1)\}=\mathcal{L}\cap\left(\mathbb{R}^2\setminus\{(0,1)\}\right)\]

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

\varnothing

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

    \[\big(\psi\circ\pi\big)(x,y)=x\]

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}.

\varnothing


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.

\varnothing

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.

\varnothing


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.

\varnothing

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.


Conclusion

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.

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