How to prove that $(0,1]$ is equinumerous to $(0,1)$

Question

So obviously the only difference between them is that $(0,1]$ includes the $1$. So my function so far is

$$ f(x) = \left\{ \begin{array}{ll} 1 & \quad ? \\ x & \quad x \neq 1 \end{array} \right. $$

I am not sure how to account for $1$ since everything is accounted every value is mapped...

are these even equinumerous to begin with?

Thank you.

Answer 1

Hint:

Handle rational point separately from irrational point. For irrational point, send them to themselves.

Edit:

Let $f_1$ be a bijection from $(0,1) \cap \mathbb{Q} \to \mathbb{N}$ and $f_2$ be a bijection from $\mathbb{N} \to (0,1] \cap \mathbb{Q} .$

Then $g(x) = f_2 \circ f_1$ is a bijection from $(0,1) \cap \mathbb{Q} \to (0,1]\cap \mathbb{Q}.$

Let $h: (0,1) \to (0,1]$, where

$$h(x) = \begin{cases} x &, x \in (0,1) \cap \mathbb{Q}^c\\ g(x) &, x \in (0,1) \cap \mathbb{Q}. \end{cases}$$