Finding a homeomorphism between $2^{\mathbf{N}}$ and the one-point compactification of $2^{\mathbf{N}}\times\mathbf{N}$. I was solving some problems found on internet to prepare for an exam. Here I think $2^{\mathbf{N}}$ is meant to be a Cantor space.
Since the $2^{\mathbf{N}}$ and the one-point compactification of $2^{\mathbf{N}}\times\mathbf{N}$ is compact and Hausdorff, I know that
a continuous bijection between them would be a homeomorphism. But I'm not sure what to do from here. Please let me know.
 A: A better way to think about this is to look for an embedding of $2^\mathbb{N}\times\mathbb{N}$ into $2^\mathbb{N}$ which misses a single point. This is a neat trick, and basically comes from "recursing" the obvious homeomorphism between $2^\mathbb{N}$ and $2^\mathbb{N}\times\mathbb{N}$. 
HINT: for $n$ fixed, think about the map $M_n$ given by 
$$
f\mapsto \overbrace{0\cdot\cdot\cdot0}^{n \; 0\text{'s}} 1 f. 
$$ That is, $M_n$ takes in an infinite binary sequence $f$, and appends $n$ $0$s and a $1$ to the beginning. So, for instance, if $$f=010000000000 . . .$$ (that is, $f$ has a $1$ in the second place, and every other bit is $0$), then $$M_3(f)=000101000000000000 . . .$$
(We allow $n=0$, in which case $M_0$ is just the map $f\mapsto 1f$.) Think about how to combine the $M_n$s to get a map $M: 2^\mathbb{N}\times\mathbb{N}\rightarrow 2^\mathbb{N}$ . . .
Now do you see why this embedding misses a single point (what is it)? Do you see how to turn this into a homeomorphism between $2^\mathbb{N}$ and the one-point compactification of $2^\mathbb{N}\times\mathbb{N}$?
A: If you already know that $2^{\Bbb N}$ is homeomorphic to the middle-thirds Cantor set $C$, you can see it as follows. For $n\in\Bbb N$ let
$$C_n=C\cap\left[\frac2{3^{n+1}},\frac1{3^n}\right]\;.$$
each $C_n$ is homeomorphic to $C$ (indeed even geometrically similar), and 
$$C=\{0\}\cup\bigcup_{n\in\Bbb N}C_n\;.\tag{1}$$
$\bigcup_{n\in\Bbb N}C_n$ is easily seen to be homeomorphic to $C\times\Bbb N$ and $C$, with the decomposition in $(1)$, to the one-point compactification of $C\times\Bbb N$.
(Note that this is really just a different version of the idea used in Noah Schweber’s answer.)
By the way, it turns out that every non-empty open subset of $C$ is homeomorphic either to $C$ or to $C\setminus\{0\}\cong C\times\Bbb N$.
A: Let $C$ be the Cantor set. For $n\in N$ take $a_n<b_n<a_{n+1}$ with $a_0=0$ and $\sup_{n\in N}b_n=1.$ For $(x,n)\in C\times N$ let $f((x,n))=a_n+(b_n-a_n)x.$ Let $S$ be the image of $f.$ Show that $f:C\times N\to S$ is a homeomorphism. Now $T=S\cup \{1\}$ is a $1$-point compactification of $S.$ Show that $T$ is homeomorphic to $C.$  
