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Let $X:= \left \{\frac{1}{n}: n\in\mathbb{N}\right \}$. Prove that there is a bijection between $[0,1]$ and $[0,1]-X$.

I still don't find a bijective function between $[0,1]$ and $[0,1]-X$. So far, I already know that $[0,1]$ and $(0,1)$ are equivalent sets. We may find a bijective function between $(0,1)$ and $[0,1]-X$ and use the composition of bijective functions.

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    $\begingroup$ Here's a hint. There's a classic way to find a bijection $f:[0, 1]\to (0, 1)$ which is to pick an infinite sequence $(s_n)$ in $(0, 1)$ and define $f(0)=s_1, f(1)=s_2, f(s_k)=s_{k+2}$, and $f(x)=x$ for $x \neq 0, 1, s_k$. Can you see how to adapt that to the case where there are infinitely many points missing? $\endgroup$ Commented Mar 6 at 15:13
  • $\begingroup$ I already have bijective function $f : [0,1]\to (0,1)$, but my question between $[0,1]$ and $[0,1]-X$. $\endgroup$ Commented Mar 6 at 15:15
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    $\begingroup$ yes, and I'm suggesting you adapt that method to the new case. $\endgroup$ Commented Mar 6 at 15:16

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The identity function is an injection $[0,1]\setminus X\to [0,1]$. If we can also find an injection $[0,1]\to [0,1]\setminus X$, we know that there is a bijection by the Schroeder-Bernstein Theorem. We can construct such an injection by sending $0$ to $0$ and "compressing" each interval $\left(\frac{1}{n+1},\frac{1}{n}\right]$ into $\left(\frac{1}{n+1},\frac{2n+1}{2n(n+1)}\right]$. This is expressed as follows:

$$f(x) = \begin{cases}0 & x= 0\\ \frac{1}{n+1}+\frac{1}{2}\left(x - \frac{1}{n+1}\right)& x\in\left(\frac{1}{n+1},\frac{1}{n}\right],n\in\mathbb{N}\end{cases}$$

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  • $\begingroup$ Is there any motivation for how to construct the function?._. $\endgroup$ Commented Mar 6 at 15:54
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    $\begingroup$ The proof of the theorem given on Wikipedia is constructive. The example there shows how to construct a bijection from $[0,1)$ to $[0,1]$; the function for this problem would be similar.0 $\endgroup$ Commented Mar 6 at 15:56
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Here's a different way to do this that works for any countable set $X$. Enumerate $X$ as $\{a_1, a_2, \ldots\}$ (meaning that $i\mapsto a_i$ should be a bijection) and choose a sequence $b_1, b_2, \ldots$ of distinct elements of $[0, 1]\setminus X$. You can do that because $[0, 1]\setminus X$ is still infinite.

Define a function $f: [0, 1] \to [0, 1]\setminus X$ by $f(a_i) = b_{2i}, f(b_i) = b_{2i-1}$, and $f(x)=x$ if $x \neq a_i, b_i$. It is pretty clearly onto (the image obviously contains everything aside from the $b_i$ because of the third case of the definition, and the first two cases show that it includes every $b_i$) and 1-1, so $f$ is a bijection.

The value of this method is that it generalises. You can replace $[0, 1]$ with any infinite set $Y$ and $X$ with any countable set such that $Y\setminus X$ is infinite. The same argument still works.

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