Is there a bijection between [0, 0.5] and (0.5, 1]?

Question

An approach I tried was to define the bijective function f:[0, 0.5] -> (0.5, 1] as,

f(x) = x + 0.5

and although this is surjective, it being injective makes the choice of adding 0.5 to x in the function seem arbitrary.

I understand that the way I have constructed the bijection is probably wrong, but I cannot figure out what the right way is.

See for example https://math.stackexchange.com/q/1233238/42969 — Martin R, Jun 05 '24 at 04:52
Does this answer your question? Proving intervals are equinumerous to $\mathbb R$ The reference from Martin R is a better duplicate and has links to more. — Ross Millikan, Jun 05 '24 at 05:05
The bijection can't possibly be continuous. The set $[0, 0.5]$ is compact and the continuous image of a compact set is continuous. Since $(0.5, 1]$ is not compact, no continuous map can be surjective. — Robert Shore, Jun 05 '24 at 06:28

score 1 · Accepted Answer · answered Jun 05 '24 at 04:57

Not an elegant bijection, but the simplest one to think about: $f:[0,0.5] \rightarrow (0.5,1]$ is defined as

$$f=\left\{\begin{split} x+0.5 &\quad x\not\in\mathbb{Q} \\ L_2(i) &\quad x\in\mathbb{Q}, x=L_1(i) \end{split} \right.$$

where $L_1$ is the sequence of all rational numbers in $[0,0.5]$ and $L_2$ in $(0.5, 1]$

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