A specific example of a CW complex and a few questions concerning it. The question I am facing is this one:
Construct a CW complex X with a 0-cell x(n) for each natural number $n \geq 0$ and a 1-cell $D_{n}^1, n \geq 1$ which is glued to $x(0)$ at one end and $x(n)$ at the other.  For each natural number $n \geq 1$ consider the segment:
$$I_n = \{t*e^{2\pi i/n }, 0 \leq t \leq 1 \}\space \mathbb{C} \cong \mathbb{R^2}$$
which has boundary points $0$ and $e^{2\pi i/n }$  From these we form the space $Y = \bigcup_{n \geq 1} I_{n} \subseteq \mathbb{C} \cong \mathbb{R^2}$ endowed with the subspace topology.
1) Give a sketch proof of the fact that Y is a closed subspace of $\mathbb{C}$. Thus $Y$ is a compact space.
2) Construct the obvious map which sends $\psi:X \rightarrow Y = \bigcup_{n \geq 1} I_{n}$ which sends $x(0)$ to the origin $0 \in \mathbb{C}$ and $x(n) \in e^{2\pi i/n }$
3) Show that $\psi$ is not a homeomorphism.
Now I am not really sure how to go about doing these, but for 1) I know that for Y to be 
closed, that all the $I_n$'swould have to be closed.  Now does this directly follow from $0 \leq t \leq 1$ as this is a closed interval?
for 2) are they just wanting 
$0, (n = 0)$
$e^{2\pi i/n}, n > 0$     ?
I feel like there has to be something more than this.. especially as t is removed, but could we do something like this: $e^{(2\pi it/n)} -1$ as we recover $0$ if $t = 0$ and $e^{2\pi i/n}$ otherwise?  Am I supposed to keep the $t$?
3) If we take the differential $\frac{de^{2\pi it/n}-1}{dt} = \frac{2\pi i}{n} e^{2\pi it/n}-1$ 
Then we have a problem at n = 0.  Not sure how to prove that it is not homeomorphic though.
Anyways, I might be way off on all 3 thoughts for the 3 parts, but any help would be truely appreciated.  Note:  This is not homework that I have to turn in.
Thanks,
Brian
 A: *

*We will show that the complement is open. Note that your space is inside the closed ball $\bar B(0,1)$. Take any point outside the ball. It clearly has a neighbourhood disjoint with $Y$ (enough to take a neighbourhood disjoint with $\bar B(0,1)$). Now take any point $x \in \bar B(0,1)\setminus Y$. In polar coordinates it is given by $(\phi,r)$ for some $(\phi,r)\in (0,2\pi)\times (0,1]$. There is a unique $n$ such that $\frac{2\pi}{n+1}<\phi<\frac{2\pi}{n}$. Now for appropriately small $r$ the open ball $B(x,r)$ is disjoint with $Y$ (easy and very formal details left to the reader).

*They want $x(0)\mapsto 0$ and $x(n)\mapsto e^{2\pi i/n}$, $D^1_n \overset{f_n}{\to} I_n$, where $n>0$. More precisely if $D_n^1=[0,1]$, then the map $f_n$ is $f_n(t) = t\cdot e^{2\pi i/n}$.

*The inverse function is not continuous. The sequence $y_n=e^{2\pi i/n}$ approaches $y_1=e^{2\pi i}=1$, but $\psi^{-1}(y_n)=x(n)$ doesn't approach $\psi^{-1}(y_1)=x(1)$ (you can easily find an open neighbourhood of $x(1)$ disjoint with other $x(n)$ - such a neigbourhood is for example the image of $(0,1] \subseteq D_1^1$).
3,5. Bonus. Actually the topology of $X$ and $Y$ differ also at $x(0)$ and $0$ ($\psi^{-1}$ is not continuous at $0$). Imagine the following open neighbourhood of $x(0)$:
$N=\bigcup_{n=1}^\infty [0,1/n)_n$, where $[0,1/n)_n\subseteq [0,1]_n=D_n^1$ (it is open as its inverse image is open in every cell). Its image via $\psi$ (inverse image via $\psi^{-1}$) is not a neighbourhood of $0$: the sequence $z_n=\frac{e^{2\pi i/n}}{n}$ obviously approaches $0$ (in polar coordinates the radius tends to $0$), but it is disjoint with $\psi(N)$ (sequence $a_n$ has limit $a_0$ if and only if for each neighbourhood of $a_0$ there are only finitely many $a_n$ outside it). The same with different words: there is no ball around $0$ in $Y$ that could be included in $\psi(N)$.
