Topology on the simplicial complex consisting of the edges $\{1,n\}$ for $\ n \in \mathbb{Z}$ Let $X$ be the simplicial complex consisting of the edges $\{1,n\}$ for $\ n \in \mathbb{Z}$, and $Y=\{re^{in}: 0\leq r\leq 1, n\in \mathbb{Z}\}$.
These 2 spaces look similar but I guess they are not homeomorphic.
I was trying to take a sequence in $Y$ converging to $0$ whose image does not converge in $X$.
Could any body explain the topology on $X$ and why it is (or not) homeomorphic to $Y$? 
Any help will be appreciated.
 A: I'm going to assume you aren't allowing the edge $\{1,1\}$ which would be a loop and makes it much easier to see that the two aren't homeomorphic.
Intuitively, the sub-space topology of Y sort of "smashes" infinitely many of the line segments together, but the simplicial topology of X allows each line segment $\{(1-t)\{1\}+t\{n\}: t>0\}$ to be separated.
In order to explicitly show that we don't have a homeomorphism, consider the map $\triangle_{1,2} : [0,1] \to X$ corresponding to the edge $\{1,2\}$. Then the set $U = \triangle_{1,2} (\{1/4 < t< 3/4\})$ is open in $X$.
Now any homeomorphism $h: X \to Y$ must map each line segment onto to a line segment in a one to one manner. It is easiest to consider the case that the image of $h$ on the edge for $\{1,2\}$ is $\{(r,0): 0\leq r \leq 1\}\subset \mathbb R^2$. All other cases are similar.
By the density of the line segments in $Y$, there are an infinite number of line segments in $Y$ that pass through any neighborhood of $h(U)$. So, by the subset topology $h(U)$ can not be open in Y.
Therefore, $X$ and $Y$ are not homeomorphic.
Edit: Intuitively, I believe that if you want to realize the topology of the simplicial complex for $\{1,n\}, n\neq 1$ as a sub-space of something familiar, then I believe that it is homeomorphic to $\cup_i \{te_i : 0\leq t\leq 1\} \subset l^2(\mathbb N)$. However, I haven't thought of it too carefully and haven't rigorously proven it to myself.
