Specify a bijection from [0,1] to (0,1].
Solution 1
$$f(x) = \begin{cases} \frac{n+1}{n+2},&\text{if $x$ has the form $\frac n{n+1}$ for $n\in\Bbb N$} \\ x,&\text{otherwise.} \end{cases}$$
That is, $f$ maps $0\to\frac12\to\frac23\to\frac34\to\ldots$ and is the identity function elsewhere in $[0,1]$.
More generally, let $\{s_1, s_2, s_3, \ldots\}$ be any countably infinite subset of $(0,1]$ and then define $$f(x) = \begin{cases} s_1 & \text{if $x = 0$} \\ s_{i+1} & \text{ if $x = s_i$ for some $i$ } \\ x & \text{otherwise} \end{cases} $$
Solution 2
See the first page of http://www.math.montana.edu/~geyer/2008/fall/361/key3.pdf for an explicit construction of the bijection and some more.
Added, so that the answer is not depended on an external source.
A bijection $f\colon (0,1) \to \mathbb R$ is given in this text as: $$f(x) = \begin{cases} 2-\frac1x & \text{for }0<x<\frac12, \\ \frac1{1-x}-2 & \text{for }\frac12\le x<1. \end{cases} $$ The inverse is $$f^{-1}(x)= \begin{cases} \frac1{2-y} & \text{for }y<0, \\ 1-\frac1{2+y} & \text{for }y\ge0. \end{cases} $$
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user108441
Updated on August 01, 2022Comments
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user108441 over 1 year
A) Specify a bijection from [0,1] to (0,1]. This shows that |[0,1]| = |(0,1]|
B) The Cantor-Bernstein-Schroeder (CBS) theorem says that if there's an injection from A to B and an injection from B to A, then there's a bijection from A to B (ie, |A| = |B|). Use this to come to again show that |[0;1]| = |(0;1]|
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MJD almost 10 years
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Martin Sleziak almost 9 yearsThe bijection in the file you link to (and other similar maps) have nice additional property that it maps rational numbers to rational numbers and irrational numbers to irrational numbers.