Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.