Let $\mathbb{R}/{\sim}$ be the quotient space given by the equivalence relation $a \sim b$ if $a$ and $b$ are rational. I am trying to understand general properties of the quotient topology and this example seems worth fleshing out in full. It's also a very strange example to me so I'd appreciate feedback on what I've figured out so far.

  • In order to figure out what the topology on $\mathbb{R}/{\sim}$ looks like we need to examine where the surjection $\pi: \mathbb{R} \to \mathbb{R}/{\sim}$ sends open sets in $\mathbb{R}$. Now any open interval $U \subset \mathbb{R}$ contains both irrational and rational points; the rationals all get sent to the same point $q$ while the irrationals get sent to separate points. So an open set in $\mathbb{R}/{\sim}$ is similar to an open set in the irrationals as a subspace of the reals (with the caveat that all open sets in $\mathbb{R}/{\sim}$ share the rational point $q$).

  • Is this space connected? I believe so as I can't think of a proper separation.

  • As Alex notes below this is not correct: My professor also mentioned this space is an example where a compact subset, namely the irrationals, is not closed. As for compactness I think it is for this reason: the rationals are dense in $\mathbb{R}$, so if we put an open neighborhood around each rational then we will cover $\mathbb{R}$. Similarly, if we put an open neighborhood around the rational point $q \in \mathbb{R}/{\sim}$, then this single neighborhood will contain all irrational points and thus be a finite cover of $\mathbb{R}/{\sim}$.

  • Are there any other significant properties of this space I should know about? In particular, is it homeomorphic to anything notable? Does it serve as a useful counterexample for any other important properties? And does this particular topology have a name?

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    $\begingroup$ You write "the rationals are dense in $\mathbb{R}$, so if we put an open neighborhood around each rational then we will cover $\mathbb{R}$." This is not true: there are open sets containing $\mathbb{Q}$ of arbitrarily small positive measure. To see this, enumerate the rationals as $\{q_n\mid n\in\mathbb{N}\}$, put an open interval of length $\epsilon\cdot 2^{-n}$ around $q_n$, and take the union. $\endgroup$ – Alex Kruckman Feb 2 '16 at 23:56
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    $\begingroup$ The density of $\mathbb{Q}$ gets you that every closed set containing $\mathbb{Q}$ is all of $\mathbb{R}$. $\endgroup$ – Alex Kruckman Feb 2 '16 at 23:57
  • $\begingroup$ Actually, 1. You write "I can see why the irrationals are open". The set of irrationals is not closed in $\mathbb{R}/\sim$, but it's also not open. Any nonempty open set in this space contains $q$. 2. The set of irrationals is not compact in this space. Probably what your professor meant is that the singleton $\{q\}$ (the "rationals") is a compact set (because it's finite) that's not closed (its complement, the irrationals, is not open). $\endgroup$ – Alex Kruckman Feb 3 '16 at 0:04

It's easy to see that the space is connected. Since every non-empty open set in $\Bbb R$ contains rationals, there is a point which meets every open set of $\Bbb R/{\sim}$. So if $U\cap V=\varnothing$ are open sets, at least one has to be empty.

Next, note that not being closed does not mean being open. The irrational numbers are not closed since they are still dense in the space, but they are not open either because the rational "point" is not closed (how could it be closed? It's dense!).

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    $\begingroup$ @AsafKargila: any continuous image of a connected space is connected, so you don't need a special argument for the quotient map in this particular case. $\endgroup$ – Rob Arthan Feb 2 '16 at 23:53
  • $\begingroup$ You're probably right, but this does offer another way of looking at the space. And it's not that the proof is particularly long. $\endgroup$ – Asaf Karagila Feb 2 '16 at 23:55
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    $\begingroup$ @AsafKargila: that's a fine response to my comment apart from "probably"! $\endgroup$ – Rob Arthan Feb 2 '16 at 23:57
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    $\begingroup$ Probably you're right about that as well! :) $\endgroup$ – Asaf Karagila Feb 2 '16 at 23:59
  • $\begingroup$ Yes. And with probability $1$ $\ddot{\smile}$. $\endgroup$ – Rob Arthan Feb 3 '16 at 0:06

I'll add my own answer, elaborating on my comments to the original post and to Rob Arthan's (now deleted) answer, just to clear things up.

A subset of $\mathbb{R}/{\sim}$ is open if and only if its preimage under the map $\mathbb{R} \to \mathbb{R}/{\sim}$ is open. Since every nonempty open set in $\mathbb{R}$ contains a rational number, the open sets of $\mathbb{R}/{\sim}$ are in bijection with $\{U\subseteq \mathbb{R}\mid U\text{ is open, and }\mathbb{Q}\subset U\}\cup \{\emptyset\}$.

How can we find open sets in $\mathbb{R}$ that contain $\mathbb{Q}$? Well, such a set is exactly the complement of a closed set which is disjoint from $\mathbb{Q}$, i.e. a closed set of irrationals. So we can take $\mathbb{R}$ and remove a single irrational, or finitely many irrationals, or a sequence of irrationals limiting to some irrational, etc. etc.

The somewhat unintuitive thing is that there are some very large (in the sense of Lebesgue measure) closed sets of irrationals. For example, fix $\epsilon>0$, enumerate the rationals as $\{q_n\mid n\in\mathbb{N}\}$, and let $O_n$ be an open interval of length $\epsilon \cdot 2^{-n}$ containing $q_n$. Then let $U = \bigcup_{n\in\mathbb{N}} O_n$. Then $\lambda(U) \leq 2\epsilon$, where $\lambda$ is the Lebesgue measure.

To see that the image of the set $I$ of irrationals in $\mathbb{R}/{\sim}$ is not compact, let $U$ be the set defined above for $\epsilon <\frac{1}{2}$ (so $U$ doesn't contain any interval $(n,n+1)$). Then for $n\in \mathbb{Z}$, let $U_n = U \cup (n,n+1)$. Now each $U_n$ is open, and $I\subseteq \bigcup_{n\in\mathbb{Z}} U_n$, but this open cover has no finite subcover. All the $U_n$ contain $\mathbb{Q}$ so the same is true of their images in $\mathbb{R}/{\sim}$.


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