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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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    @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.2012-08-14

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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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    @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".2015-10-09
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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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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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    But what topology we get for the quotient space $\mathbb{R}/\mathbb{Z}$? If the equivalence relation is $x\tilde~ y$ iff $x-y\in\mathbb{Z}$?2016-03-15
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The quotient group R/Q is similar to R/Z in some respects, but is quite different and, I think, impossible to visualize in the way R/Z is. First note that if p is a rational number, then it's equivalence class (i.e. coset generated by p) in R/Q, denoted [p] is equal to [0]. That is, all rationals collapse to the single coset Q. Now, note that if r is an irrational number, we can write it as r=n+s, where s is an irrational number in the interval (0,1) and n is an integer. That means that r-s=nis a rational number, which in turn means r and s are in the same equivalence class, i.e. [r] = [s]. That means the elements of R/Q look like {0} U {a set of irrationals in the interval (0,1)}. But what set of irrationals, exactly? Not all of them; for example take the decimal portion of PI (=0.14159...) and add 0.5 to it to get 0.64159.... Both are irrational numbers yet their difference is ½, so they generate the same coset, in other words, they are collapsed to the same element of R/Q. On the other hand, it is known that sqrt(2), sqrt(3) and sqrt(2)-sqrt(3) are all irrational numbers. That means the cosets [sqrt(2)] and [sqrt(3)] are distinct elements of R/Q. So, some irrationals collapse to the same element in R/Q but not all do. So the question becomes: is there a way to choose or describe a set of irrational numbers that represent the distinct non-zero cosets of R/Q? The Axiom of Choice implies that, yes, one can choose a set of irrational numbers in the interval (0,1) that form a distinct complete set of cosets for R/Q. The catch, however, is that the Axiom of Choice gives no recipe for how to choose or describe such a set of representatives. What we can say is that two non-zero elements of R/Q, call them [r] and [s], are equivalent if and only if their decimal representations differ by only a finite number of digits. So a non-zero coset consists of all the irrational numbers in (0,1) that differ from each other by only a finite number of digits. It would seem that surely there must be a way to methodically pick from each coset a "canonical" representative. A likely candidate might be to pick the smallest member in each coset, but of course that fails because there is no smallest member in each non-zero coset; same for largest. What one would ideally like is a choice function f:R/Q -> (0,1) such that for any two cosets C and D, f(C+D) = f( C)+f(D) mod(0,1). As far as I know, no such choice function has been described and, indeed I do not know if it is even possible to define such a function in the standard ZF language. Note that the AC does not imply that such a function as f, above, exists. It only says that a choice function exists, but says nothing about how it will behave arithmetically as described above.

(*)Note: The statement above that two cosets [r] and [s] are equal iff r and s differ by a finite number of digits is almost right, but ignores the possibility that r-s might be a repeating decimal, like 1/9. If we permit ourselves to use the repeating decimal symbology of a bar over the repeating segment of decimals, then the original statement stands true.

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    The "block of text" approach makes it hard to "visualize" what new information you may be adding with this late answer (the Question was posted closed to five years ago). If you could use paragraphs and highlight in some way the added material, it will make it easier for Readers to appreciate.2017-05-21