Specify a bijection from [0,1] to (0,1].

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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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Updated on August 01, 2022

Comments

  • user108441
    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]|

  • Martin Sleziak
    Martin Sleziak almost 9 years
    The 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.