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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 $f(t) = e^{2\pi i t}$?

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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
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

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.

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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
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. – Grumpy Parsnip Feb 26 '13 at 22:16

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.

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See comments on the question for why this is not perfect. – Dan Rust Feb 26 '13 at 21:53
Edited. $\,\,\!\!\,$ – Berci Feb 26 '13 at 21:55

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