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I was wondering about this. I know it is possible to visualize the quotient group $\mathbb{R}/\mathbb{Z}$ as a circle, and if you consider these as "topological groups", then this group (not topological) quotient is topologically equivalent to a circle.

But then, what does $\mathbb{R}/\mathbb{Q}$ look like?

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It looks like a mess... I doubt there is anything more useful you can say about it. – Mariano Suárez-Alvarez Aug 13 '12 at 23:25
It looks like a point, since with the quotient topology, every continuous function $\hspace{1.6 in}$ from $\:\mathbb{R}/\mathbb{Q}\:$ to a Hausdorff space is constant. $\;\;$ – Ricky Demer Aug 13 '12 at 23:28
@MarianoSuárez-Alvarez: I believe it's actually isomorphic to $\mathbf R$ (as a group or $\mathbf Q$-vector space), since $\mathbf R$ is actually a direct sum of $\mathfrak c$ copies of $\mathbf Q$ (obviously, since it is a vector space over rationals), and we're taking the quotient by one of the copies in the sum (or a one-dimensional subspace). So not that much of a mess. – tomasz Aug 14 '12 at 0:20
@tomasz: But infinitely generated vector spaces over the rationals, or worse: $\mathbb Z$-modules, give you no intuition on how something "looks like". The rationals themselves look like a mess. They are totally disconnected without isolated points, and totally disconnected spaces with no isolated points are messy in terms of visualization. Approaching this as a vector space is in fact a rather formal way of looking at this. – Asaf Karagila Aug 14 '12 at 0:57
@tomasz, indeed, that is clear (every infinite dimensional vector space is isomorphic to each of its quotients over a finite dimensional subspacees) but here we are looking at the quotient as a topological group. – Mariano Suárez-Alvarez Aug 14 '12 at 1:42
up vote 19 down vote accepted

So, you say that the group (not topological) quotient of $\mathbb{R}/\mathbb{Z}$ is topologically equivalent (i.e., homeomorphic) to the circle. However, this doesn't make any sense unless you have a topology on $\mathbb{R}/\mathbb{Z}$! More the point is that a topological group like $\mathbb{R}$ has both a topological structure and a group structure. Now, when you form the group quotient $\mathbb{R}/\mathbb{Z}$, it can be given a topological space in a natural way, in particular, via the quotient topology. Notice that when we do this we again get a topological group (i.e., the quotient group operations are continuous with respect to the quotient topology). Furthermore, the quotient $\mathbb{R}/\mathbb{Z}$ (as a topological space) is homeomorphic to the circle.

Now, in the case of your question, the quotient topology on $\mathbb{R}/\mathbb{Q}$ is the trivial topology. This is not hard to prove since preimages of open sets must be open and saturated. Thus if such a preimage is nonempty, it contains an open interval, and since it is saturated, it must contain all real numbers which differ by a rational from a point in this interval. It is then easy to see that this set must be all of $\mathbb{R}$. Thus the only saturated open sets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$ itself. Hence the quotient topology is trivial. Furthermore, it is trivial that any map into a space with the trivial topology is continuous, so the quotient group operations on $\mathbb{R}/\mathbb{Q}$ are again continuous. So we again have a topological group, albeit not a very interesting one because it isn't very interesting as a topological space. As far as what this space "looks" like, it is similar to a one point space for the reason Ricky mentioned in the comments. However, it is not really easy to visualize since it is not homeomorphic to any subspace of $\mathbb{R}^n$ equipped with the subspace topology (because it is not Hausdorff, or any one of a number of other reasons).

Edit: I should have added that whenever you have a topological group and form the quotient in the way we did above the result is always a topological group. However, unless the original normal subgroup is closed, the resulting quotient group will not even be $T_0$ as a topological space. Thus it is only really interesting to form the quotient when the set by which you quotient out is closed. This explains why $\mathbb{R}/\mathbb{Z}$ is interesting as a topological group, but $\mathbb{R}/\mathbb{Q}$ is not.

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What I meant by "group quotient, not topological quotient" didn't mean there was no topology, rather it meant that when considering "$\mathbb{R}/\mathbb{Z}$" as a topological space, you use the topology of $\mathbb{R}/\sim$ with $\sim$ being the coset-forming equivalence relation, as opposed to the relation given by $a \sim b \iff a, b \in \mathbb{Z}$. – mike4ty4 Aug 14 '12 at 3:06
@mike4ty4: I assumed that was what you were getting at, it just wasn't clear to me when I originally read the question. My apologies. – J. Loreaux Aug 14 '12 at 23:03
Coming back to this, would a good way to think of it be "it is like a point, but it is actually all points of $\mathbb{R}$, or an $\mathbb{R}$-sized set, all 'mushed' and glued to each other together in that 'point'"? IOW, it looks like a point but it's actually a nasty, freaky and scary thing with a whole continuum of points all mashed on top of each other. – mike4ty4 Oct 9 '15 at 8:05
@mike4ty4: It really depends on which aspect of $\mathbb{R} / \mathbb{Q}$ you care about. From the topological standpoint, it is indistinguishable from any other set with the trivial topological and cardinality of the continuum (because they are homeomorphic), and it is even indistinguishable from a one-point space when the codomain is Hausdorff. However, from the group standpoint, this is a large object which is (group) isomorphic to a $\mathfrak{c}$-dimensional vector space over $\mathbb{Q}$. So, the answer to your question is simultaneously "yes" and "no". – J. Loreaux Oct 9 '15 at 11:54

It really depends on what you think about as visualizing.

The group $\mathbb Z$ is discrete, so between two successive points there is a part which looks a bit like $\mathbb R$. The result, if so, is somewhat close to being $\mathbb R$.

On the other hand, $\mathbb Q$ is a dense subgroup of $\mathbb R$. This means that it gets a lot messier. Not without a good reason too, we can usually imagine things which have shape, things which can be measured.

Any set of representatives for $\mathbb R/\mathbb Q$ cannot be measured. This tells you that it is practically impossible to visualize this quotient in the same sense that we would imagine a circle, a ball, or even if we try really hard and we imagine a four-dimensional space.

Furthermore, using the axiom of choice we can create such set of representatives; however without the axiom of choice this quotient might not even be linearly ordered. Namely, it forms a set which cannot be linearly ordered. In contrast, $\mathbb R/\mathbb Z$ is a circle, or a half-open interval (where we identify the endpoints), even without the axiom of choice.

This tells you even more: you need the axiom of choice to impose an order on this set. Just a linear order, not even a well-order. Therefore imagining this as a linearly ordered set is even harder than we may believe at first.

My suggestion is not to try and visualize it. Accept this as a formal object which you can understand to some extent, but not see. Move on with this. Eventually, after running into infinitary objects ($\ell^2$, for example) and succeeding in visualizing those -- come back to this one, then you might be able to pull this off.

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In my long but not very broad experience, $\mathbb{R}/\mathbb{Q}$ is not a natural object: it never came up anywhere near my own work. So I would consider it a curiosity, and in some sense not worthy of our trying to visualize it. – Lubin Aug 14 '12 at 1:12
@Lubin: That would immensely depend on what your experience is. Most people could say that sets of cardinality larger than $2^\frak c$ are not naturally occurring, but in set theory they occur all the time. – Asaf Karagila Aug 14 '12 at 1:13
I found this: I'm a little amazed that what at first looks like an innocuous quotient is such a crazy, pathological, and bizarre entity. – mike4ty4 Aug 14 '12 at 5:29
But what topology we get for the quotient space $\mathbb{R}/\mathbb{Z}$? If the equivalence relation is $x\tilde~ y$ iff $x-y\in\mathbb{Z}$? – Zzz Mar 15 at 19:44

If you ignore topology, it's pretty much the same as $\mathbf R$.

Notice that $\mathbf R$ is a $\mathfrak c$-dimensional vector space over $\mathbf Q$, of which $\bf Q$ is a one-dimensional subspace. Taking the quotient $\bf R/\bf Q$ is actually taking the quotient of a $\mathfrak c$-dimensional vector space by a one-dimensional subspace, which is again a vector space, and is still $\mathfrak c$-dimensional (because $1<\mathfrak c$ ;) ), so it is isomorphic to $\bf R$ as a vector space over $\bf Q$, and in particular as a group.

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