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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
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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
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up vote 16 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
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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
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I found this: en.wikipedia.org/wiki/Vitali_set 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
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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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