Prove $|(0, 1]| = |\mathbb{R}|$ I could find proof for $|(0,1)| = |\mathbb{R}|$, but could not find anything about $|(0, 1]| = |\mathbb{R}|$. I do not know how to prove this. The only way I can think of is showing that there is a bijection from $(0, 1] \rightarrow (0,1)$, then we will have $|(0, 1]| = |(0,1)| = |\mathbb{R}|$. However, I would like to prove this by using The Schroder-Bernstein theorem with $(0,1]$. What would be the key difference between $|(0,1)| = |\mathbb{R}|$ and $|(0, 1]| = |\mathbb{R}|$? I think there still should be the bijection from $(0,1]$ to $\mathbb{R}$, but I'm struggling with coming up with idea.
 A: Lemma: If $X$ is an infinite set and $a \not \in X$ then $|X \cup \{a\}| = |X|$.
SO if $|(0,1)| = |\mathbb R|$ then $|(0,1]|=|(0,1)\cup \{1\}| = |(0,1)| = |\mathbb R|$.
SO let's prove the lemma.
Let $a \not \in X$ and $X$ is not empty so there is an $x_1\in X$.  Let $\phi(a)= x_1$.
$X$ is infinite so $X_1= X\setminus \{x_1\}$ is infinite (why?) so there is an $x_2\in X_1$.  So let $\phi(x_1) = x_2$.
Via recursion induction we can define $X_{k} = X_{k-1} - \{x_{k-1}\}$ is infinite and and $x_{k} \in X_{k}$ as $\phi(x_{k-1}) = x_k$.
So we end up with an infinite countable set $\{a, x_1, x_2, .....\}$ with $\phi(a) = x_1$ and $\phi(x_k) = x_{k+1}$.  For all $w \in X \setminus \{a, x_1, x_2, .....\}$ we can just let $\phi(w) = w$.
That is a bijection.
.....
Or a practical example.  Let $\phi(1) = \frac 12$.  And $\phi (\frac 1{2^k}) = \frac 1{2^{k+1}}$.  Other wise let $\phi(w) = w$.
That is a bijection of $(0,1]\to (0,1)$.  Compose it with your bijection of $(0,1)\to \mathbb R$.
That is.... if $\chi: (0,1)\to \mathbb R$ then let $\phi(\frac 1{2^k}) =\chi (\frac 1{2^{k+1}})$.  and for all other $w$ let $\phi(w) = \chi(w)$.
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Ah,  Shroeder Berstein!  That makes it easy!
Let $f:(0,1] \to \mathbb R$ be $f(x) =x$.  That's clearly injective. (but not onto)
And let $g:\mathbb R \to (0,1]$ be... well, whatever you did for $\mathbb R\to (0,1)$ extend it to $(0,1]$ and it will be injective but not unto....
We can use $\arctan: \mathbb R \to (-\frac \pi 2, \frac \pi 2)$ is bijective.
So $\frac {\arctan x +\frac \pi 2}{\pi}: \mathbb R \to (0,1)$ is bijective.
So $g(x) = \frac {\arctan x +\frac \pi 2}{\pi}$ so $g:\mathbb R \to (0,1]$ is injective.  (It's not onto as there is no $x: g(x)=1$ but we don't need it to be onto.  Just injective.)
So as we have injective  $f:(0,1]\to \mathbb R$ and injective $g:\mathbb R\to (0,1]$ we know a bijection $\chi: (0,1]\to \mathbb R$ must exist.
Just what is that bijection?.... who the @#%\$ knows and who the @#%\$ cares?  We have the SB theorem so we don't have to care.  Citing theorems is free.  Thinking will cost you extra.
