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Consider the topological spaces $A \subseteq X$. Identify the quotient space $X/A$ as a more familiar topological space and prove its homeomorphic.

$X = \mathbb{R}$ and $A = \mathbb{Z}$

My thought was that $X/A$ is homeomorphic to a circle. Is this the right idea? If it is, would I use $f(t) = e^{2\pi i t}$?

user64013
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    In topology, $X/A$ generally means the space you get from $X$ by gluing together all the elements of $A$. That is very much not a circle here. – Chris Eagle Feb 26 '13 at 21:50
  • But if you mean the quotient group with its natural quotient topology, then yes. – Chris Eagle Feb 26 '13 at 21:50
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    Careful. Normally, when $A\subset X$ and we then take the quotient $X/A$, we mean that every element of $A$ is identified to a point. What you are describing seems to be the orbitspace construction given by the integer translation action of $\mathbb{Z}$ on $\mathbb{R}$. These are NOT the same space. Confusingly they have the same notation. The space given by identifying $\mathbb{Z}$ in $\mathbb{R}$ is the countably infinite bouquet of circles. – Dan Rust Feb 26 '13 at 21:51
  • Ahhh.. I see the point – Berci Feb 26 '13 at 21:51
  • Hmm I seem to be getting mixed thoughts on this problem. Would it help to know that this is the quotient topology? – user64013 Feb 26 '13 at 21:55

2 Answers2

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My apologies for my initial comment: I was thinking about algebra, not topology.

If you identify $\Bbb Z$ to a point, the open intervals between consecutive integers remain distinct. Each interval $[n,n+1)$ is (so to speak) bent around into a circle, and all the circles have one point in common, the integer point. You get the same space if you take $\Bbb Z\times S^1$, where $S^1$ is the unit circle, fix a point $p\in S^1$, and identify the set $\Bbb Z\times\{p\}$ to a point. Think of a book with a page for each integer. Cut away all of each page except a circle tangent to the spine of the book at the centre of the spine. The resulting object is your space.

Added: If you make the circles different sizes, you can also visualize it in the plane:

enter image description here

(This is Wikipedia’s picture of the Hawai`ian earring.)

Note, however, that you have to take this visualization with a grain of salt: if you view it a subspace of the plane with the subspace topology, you find that the points on the $x$-axis converge to the origin, while the points at the centres of the intervals $[n,n+1)$ do not converge to the common point of the quotient $\Bbb R/\Bbb Z$. A better picture would have the circles expanding outward, with larger and larger radii, instead of contracting inward, but so far I’ve not found a suitable picture.

Added2: And even with the better picture, you need to be careful, because the quotient space still does not have the same topology as the subset of the plane: it is not first countable at the origin (the point corresponding to $\Bbb Z$). Any subset of $\Bbb R$ that contains an open interval around each integer yields a nbhd of the point corresponding to $\Bbb Z$ in the quotient, but not all of them yield nbhds of the origin in the subset of the plane consisting of those expanding circles. The circles are still a helpful tool for visualizing the quotient space, however.

Brian M. Scott
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  • Well, if you did that with the book pages, all the circles clumped together would make a cylinder then wouldn't it? – user64013 Feb 26 '13 at 22:02
  • @user64013: Open the book and flap it! :-) – Brian M. Scott Feb 26 '13 at 22:02
  • Ahhhhhhh. Flapping implies a sphere. At least thats what I am visually getting. – user64013 Feb 26 '13 at 22:03
  • @user64013: You should visualize it with each page sticking out at a slightly different angle, so that two pages touch only at the one common point. – Brian M. Scott Feb 26 '13 at 22:04
  • lol then I have no idea what to call that. – user64013 Feb 26 '13 at 22:06
  • You can simply take $${(x,y)\in\mathbb R^2\mid \exists n\in\mathbb N\colon (x-n)^2+y^2=n^2} $$ if that's easier than Brian's book analogy – Hagen von Eitzen Feb 26 '13 at 22:10
  • @user64013: I’ve added a planar visualization, complete with picture, which might help. – Brian M. Scott Feb 26 '13 at 22:12
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    The Hawaiian earring has a different topology than $\mathbb R/\mathbb Z$. For example, if you take the half-integers $n+1/2$, they don't converge to a point in $\mathbb R/\mathbb Z$, even though the middles of the circles converge to a point in the planar embedding above. – Cheerful Parsnip Feb 26 '13 at 22:16
  • @Jim: I know that and will be adding a cautionary comment; with this picture I’m just trying to get across the general idea. I was hoping to find a picture with expanding circles. – Brian M. Scott Feb 26 '13 at 22:18
  • I see what you mean now. Its a neat visualization. Since the question asks to identify a more familiar topological space, would I just choose one circle, and conclude it as a circle? The reason I ask this is because this question came from an introduction to quotient topology. – user64013 Feb 26 '13 at 22:24
  • @BrianM.Scott: sounds good. The Hawaiian earring is such an interesting space I couldn't bear to see it treated like a wedge of circles. ;) – Cheerful Parsnip Feb 26 '13 at 22:25
  • @user64013: I’m a bit puzzled by the wording of the question: I can’t think of any way to describe this quotient as a more familiar space. It definitely isn’t a circle, but I’m wondering whether the person who posed it had the same mental hiccup that Berci and I had initially. (And you’re very welcome.) – Brian M. Scott Feb 26 '13 at 22:26
  • Maybe thats why I was a little confused too. The question was to vague. – user64013 Feb 26 '13 at 22:30
  • I went and asked the person who made this problem up. They said I am pulling all the integers together. Does that imply the picture above then? – user64013 Feb 26 '13 at 23:08
  • @user64013: If you turn it inside-out, so to speak, and make the circles expand instead of contract. Make them all tangent at the origin, but put the centres at $\langle n,0\rangle$ instead of $\langle\frac1n,0\rangle$ as in the picture. – Brian M. Scott Feb 26 '13 at 23:14
  • Your picture is incorrect: here is a valid illustration: http://i.imgur.com/39GUbxq.png – Steven Clontz May 06 '16 at 19:20
  • @Steven: I know that it’s incorrect. I explicitly said that it’s incorrect. Yours is better, but I’d prefer one that does not give the impression of converging to a specific circle. – Brian M. Scott May 06 '16 at 19:22
  • Ah, the real issue is that it's not embeddable in the plane at all due to a non-first-countable point, making mine also invalid. It would be more clear to preface the image with that fact rather hide it in a fifth addendum; as shown a drive-by reader would be reasonably confused into thinking it's metrizable at a glance. (The excluded circle of diamater 2 isn't a problem at all, though.) – Steven Clontz May 06 '16 at 19:51
  • @Steven: When I have a bit more time I’ll probably rewrite it completely. (I may even try to find or make a better picture; I don’t like yours much better than the one that I took from Wikipedia, since I do consider the dashed circle visually misleading.) But the second addendum covers the most important point (that I actually thought I’d mentioned in the first), so it’s not right at the top of my list. – Brian M. Scott May 06 '16 at 19:58
  • Maybe we should agree to disagree, but the dashed circle is always there as part of a natural compactification of the space. – Steven Clontz May 06 '16 at 20:01
  • @Steven: That’s not clear. How do you define the topology at the origin? – Brian M. Scott May 06 '16 at 20:13
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Perfect (in case of topological groups and their quotients).

But, as Chris Eagle pointed out in the comment, purely topologically it is something else: all the points of $\Bbb Z$ are glued together to one new point but the other points remain. So, in that case, it is countably infinitely many circles glued together in one point.

Berci
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