Topology - Quotient Space and Homeomorphism 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 $f(t) = e^{2\pi i t}$? 
 A: 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:

(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.
A: 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.
