Show that $\aleph_0 + 2^{\aleph_0} = 2^{\aleph_0}$ by establishing a bijection out of $|\{1,2,3,\ldots\} \cup (0,1)| = |(0,1)|$.
Workings:
I believe that I need to pull a countably infinite number of points out of $(0,1)$ while still being able to fill in the countably infinite number of holes this would leave.
Any help will be appreciated
We can extract a countable set out of $(0,1)$, one of such example is $A_1=\{\frac1n : n\in \mathbb{N}\}.$ Then let $A_3 = (0,1) \backslash A_1$ be the elements of $(0,1)$ outside $A_1$. Also, let $A_2 = \mathbb{N} = \{1,2,3,\ldots\}$.
Then both $A_1$ and $A_2$ are countable set. So, we have a bijection
$$ g: A_1\cup A_2 \rightarrow A_1.$$
This is Hilbert's hotel argument as commented by @Dan Rust.
Now we make a bijection as: $$ f(x) = x \ \ \mathrm{if} \ \ x\in A_3 $$ $$f(x) = g(x) \ \mathrm{if} \ \ x\in A_1\cup A_2.$$ Then $f$ is a bijection between $A_1\cup A_2\cup A_3$ and $A_1\cup A_3$.
Let $A=\{2^{-n}:n\in \mathbb N\}.$ For $x\in (0,1)$ \ $A$ let $g(x)=x.$ For $x=2^{-n}\in A$ let $g(x)=2^{-2n}.$ For $m\in \mathbb N$ let $g(m)=2^{1-2m}.$ Then $g:((0,1)\cup \mathbb N)\to (0,1)$ is a bijection.
The idea is to take some $A=\{a_n: n\in \mathbb N\}\subset (0,1)$ where $m\ne n\implies a_m\ne a_n$, and map $a_n$ to $a_{2n}$, while mapping each $x\in (0,1)$ \ $A$ to itself. This leaves room, that is, $\{a_{2n-1}: n\in \mathbb N\}$ for a copy of $\mathbb N.$