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I need a bijection from $[0, 1]\cup(1,2)\cup${$3$} to $(0, 1)$.

At first I thought it was a trivial problem, but after struggling with it for some time I think it's harder than it looks.

For instance, if you try to try to send $(0, 1)$ to $(0, \frac{1}{2})$ and {$3$} to $\frac{1}{2}$ and ($1, 2$) to ($\frac{1}{2}, 1$), then you are still left with a place to send $0, 1$ since they are included in [$0, 1$].

I've tried some variations but can't get it. Help/hints would be appreciated.

Wilson Brians
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  • Do you absolutely need to show a particular bijection, or would it be enough for you to prove that one exists? – hmakholm left over Monica Nov 15 '16 at 18:46
  • I need an explicit one.@HenningMakholm – Wilson Brians Nov 15 '16 at 18:48
  • Your set $[0,2)\cup{3}$ is clearly bijective to $[0,2]$ and therefore to $[0,1]$ also (via the map $x\mapsto x/2$). This reduces your problem to construct a bijection between $[0,1]$ and $(0,1)$, which already has an answer here at MSE. –  Nov 15 '16 at 19:32
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    @mathbeing very nice. I kept trying to break down the (0, 1) but breaking it down from the other side is much nicer. But this still isn't an explicit bijection in the way that the main answer is. – Wilson Brians Nov 15 '16 at 19:51
  • @mathbeing I imagine your answer in my head like this. We have four bins full of elements. You stuff the elements of bin 1 into bin 2, then stuff everything from this bin into bin 3, then everything from this bin into bin 4. But what I need is to start off with all the elements in one bin and stuff them all into bin 4 at the same time. Mathematically I think it's quite different, no? Even though I know it still proves the important result about equivalent cardinality. – Wilson Brians Nov 15 '16 at 20:04
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    It is the same via composition of bijections; the answers given in the above link solve the problem of constructing a bijection $\varphi:[0,1]\to(0,1)$, and you can explicitly write another bijection $\phi$ from your set to $[0,1]$, and then $\varphi\circ\phi$ will be the bijection you're looking for. –  Nov 15 '16 at 20:12

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This is a common approach: $$ \begin{align} 0&\mapsto1/2\\ 1&\mapsto1/4\\ 3&\mapsto1/8\\ 1/2^n&\mapsto 1/2^{n+3}&&\text{for }n=1,2,...\\ x&\mapsto x/2&&\text{otherwise} \end{align} $$ So effectively we are fitting the countable set $\{1/2^n:n\in\mathbb N\}\cup\{0,1,3\}$ into $\{1/2^n:n\in\mathbb N\}$ by starting by mapping $0,1,3$ to the first three elements thus shifting the rest of the elements three places. Note that $x\notin\{1/2^n:n\in\mathbb N\}\cup\{1\}$ implies $x/2\notin\{1/2^n:n\in\mathbb N\}$, so this produces no collisions.

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