Showing homeomorphism between interval's quotient spaces We have two spaces: $[0,1]/C$ and $[0,1]/A$, where $C$ denotes the Cantor set and $A=\{0,1,\frac{1}{2},\frac{1}{3},...\}$. One needs to show they are homeomorphic. 
What I thought about is showing $A\cong C$, but I suppose it is not true... What should I do? Show the same sets are open in both spaces?
 A: Let $X_C=[0,1]/C$ and $X_A=[0,1]/A$. Let $p_C$ be the point of $X_C$ corresponding to $C$, and let $p_A$ be the point of $X_A$ corresponding to $A$. $[0,1]\setminus C$ is the union of a countably infinite family $\mathscr{I}_C$ of pairwise disjoint open intervals, and $X_C\setminus\{p_C\}$ is clearly homeomorphic to $[0,1]\setminus C=\bigcup\mathscr{I}_C$. Similarly, $[0,1]\setminus A$ is the union of a countably infinite family $\mathscr{I}_A$ of pairwise disjoint open intervals, and $X_A\setminus\{p_A\}$ is homeomorphic to $[0,1]\setminus A=\bigcup\mathscr{I}_A$. Thus, the subspaces $X_C\setminus\{p_C\}$ and $X_A\setminus\{p_A\}$ are homeomorphic to $(0,1)\times\Bbb N$ and hence to each other.
Let $\pi_C:[0,1]\to X_C$ and $\pi_A:[0,1]\to X_A$ be the quotient maps. 
$U\subseteq X_C$ is an open nbhd of $p_C$ iff $\pi_C^{-1}[U]$ is an open nbhd of $C$ in $[0,1]$. Let $V=\pi_C^{-1}[U]$ be an open nbhd of $C$, and let $K=[0,1]\setminus V$. $C$ and $K$ are disjoint closed sets, so 
$$d=\inf\{|x-y|:x\in C\text{ and }y\in K\}>0\;.$$
It follows that if $I\in\mathscr{I}_C$ has length less than $d$, then $I\subseteq V$. Thus, $V$ contains all but finitely many of the intervals in $\mathscr{I}_C$. Moreover, if $I=(a,b)\in\mathscr{I}_C$ is not a subset of $V$, then $V\cap I$ is open in $I$, and there are $c,d\in I$ such that $(a,c)\cup(d,b)\subseteq V$. Conversely, it’s not hard to see that any subset $V$ of $[0,1]$ that contains $C$ and all but finitely many members of $I$ and has the property described in the previous sentence, then $V$ is an open nbhd of $C$, and $\pi_C[V]$ is an open nbhd of $p_C$.
$X_C$ is therefore homeomorphic to the following space.

Let $Y=(0,1)\times\Bbb N$, let $p$ be a point not in $Y$, and let $X=\{p\}\cup Y$. $\Bbb N$ has the discrete topology and $Y$ the corresponding product topology. $U\subseteq X$ is an open nbhd of $p$ iff 
  
  
*
  
*$p\in U$,  
  
*$(0,1)\times\{n\}\subseteq U$ for all but finitely many $n\in\Bbb N$, and  
  
*if $(0,1)\times\{n\}\nsubseteq U$, then 
  
  
*
  
*$U\cap\big((0,1)\times\{n\}\big)$ is open in $(0,1)\times\{n\}$, and  
  
*there are $a,b\in(0,1)$ such that $\big((0,a)\cup(b,1)\big)\times\{n\}\subseteq U$.
  
  

I’ll leave it to you to show that $X_A$ is also homeomorphic to $X$; you can pretty much just substitute $A$ for $C$ throughout the argument for $X_C$.
