Topologist's Comb I've already posted another question because I've got an assignment to finish (still got half a day left :/) and had to realise that I'm a bit lost.
Anyway, the question has to do with the following comb space:
$Y=\{(x,0):x\in[0,1]\}\cup\{(1/n,y):y\in[0,1],n\in\mathbb{N}\}$
I have to:
a) Draw it. (which is easy...)
b) Find a bijection from $[0,1]$ to Y.
c) Show that $[0,1]$ and Y are not homeomorphic.
I'm struggling with the bijection at the moment. Any help?
Cheers
Tom
 A: Here is a suggestion for the bijection.  Take the tooth of the comb at $x=\frac{1}{n}$, shrink it to a length of $\frac{1}{2n(n+1)}$, and map it to the unit interval
$$\left[\frac{1}{2n+2}+\frac{1}{2n},\frac{1}{n}\right]$$
Then shrink the open intervals $(\frac{1}{n+1},\frac{1}{n})$ on the base of the comb to $(\frac{1}{n+1},\frac{2n+1}{2n(n+1)})$.
Visually, we're shrinking each tooth and folding them onto the base, but we shrink each enough to be able to map the base into itself such that the base's image is the complement of the image of the teeth.
Make sense?
Also, a hint for the third part of the problem: Notice that the points $(0,a)$ are limit points of $Y$ for $a\in[0,1]$, but $(0,a)\in Y$ if and only if $a=0$.
A: Consider the bijective mapping f:[0,1] to (0,1] defined by
f(a,b) = $\left\{
     \begin{array}{lr}
       \frac{1}{2} & : x=0\\
       \frac{1}{n+1} & : x=\frac{1}{n}\\
       x & : otherwise\\ 
     \end{array}
   \right.$
Now try and map the set $\{(x,0):x\in[0,1]\}$ to say [0,$\frac{1}{2}$) and then map $\{(1/n,y):y\in(0,1]\}$ for each fixed n to [$\frac{1}{n+1}$,$\frac{1}{n+2}$). This will give you a bijective function. 
I have muddled it up a little, as I am in a hurry now but I think the gist is clear.
