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How can I construct a one-to one correspondence between the Set $\left [ 0,1 \right ]\bigcup \left [ 2,3 \right ]\bigcup\left [ 4,5 \right ] ... $ and the set $\left [ 0,1 \right ]$ I know that they have the same cardinality

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You can prove they have the same cardinality by showing that you can inject each set into the other and applying the Berstein-Schroeder Theorem. The proofs of this theorem constructs a bijection using the two injections. – William Aug 31 '12 at 8:51
The reason this is not going to be an "obvious" function is that such function cannot be continuous. – Asaf Karagila Aug 31 '12 at 9:10
Adapt this answer. – Did Aug 31 '12 at 9:38
@AsafKaragila: Just out of curiosity, what argument did you have in mind that would contradict continuity from this union set to $[0,1]$? – T. Eskin Aug 31 '12 at 9:58
@ThomasE.: I had in mind the other direction. :-) But generally speaking, finding bijections which are continuous-but-not-open-as-well is not a trivial task either, especially for a beginner. – Asaf Karagila Aug 31 '12 at 10:08

2 Answers 2

up vote 10 down vote accepted

Suppose that you had a bijection $f:[0,1]\to(0,1]$. Then you could decompose $[0,1]$ as

$$\begin{align*} [0,1]&=\left[0,\frac12\right]\cup\left(\frac34,1\right]\cup\left(\frac58,\frac34\right]\cup\left(\frac9{16},\frac58\right]\cup\dots\\ &=\left[0,\frac12\right]\cup\bigcup_{n\ge 1}\left(\frac{2^n-1}{2^{n+1}},\frac{2^{n-1}+1}{2^n}\right]\;, \end{align*}$$

map $[0,1]$ to $\left[0,\frac12\right]$ in the obvious way, and for $n\ge 1$ map $[2n,2n+1]$ to $\left(\frac{2^n-1}{2^{n+1}},\frac{2^{n-1}+1}{2^n}\right]$ using straightforward modifications of $f$ for each ‘piece’. I’ll leave that part to you unless you get stuck and ask me to expand; the hard part is finding $f$. Here’s one way:

$$f:[0,1]\to(0,1]:x\mapsto\begin{cases} \frac12,&\text{if }x=0\\\\ \frac1{2^{n+1}},&\text{if }x=\frac1{2^n}\text{ for some }n\ge 1\\\\ x,&\text{otherwise}\;. \end{cases}$$

In other words, $f$ is the identity map except on the set $\displaystyle{\{0\}\cup\left\{\frac1{2^n}:n\ge 1\right\}}$, which it shifts one place ‘forward’ like this:


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+1. Just this: the final "permutes" is not a permutation, which in fact is exactly the point of the exercise. – Marc van Leeuwen Aug 31 '12 at 8:56
@Marc: Very true; thanks! – Brian M. Scott Aug 31 '12 at 9:01
Compare with this answer. – Did Aug 31 '12 at 9:38

For $x\in{(k,k+1)}$ with $k\geq 2$ and $k$ even define $f(x)=\frac{1}{2x-k}$; for $x\in(0,1]$ define $f(x)=\frac{x+1}{2}$; set $f(0)=0$. Now it remains to map $A=\{2,3,4,5,..\}$ bijectively to $\{1/2,1/4,1/6,1/8..,\}$ to do this define $f(x)=\frac{1}{2(x-1)}$ on $A$. This should give you the desired bijection. Please, let me know if I have made any silly error this time.

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I've cleaned the formulation up slightly. I see no error. You might have done a bit simpler with $\frac1{x-k/2+1}$ for $x\in{(k,k+1)}$ and any even $k$, I think. – Marc van Leeuwen Aug 31 '12 at 9:59
@Marc van Leeuwen thank you for the feedback, I appreciate it very much! – Henrik Aug 31 '12 at 10:06

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