How to construct a one-to one correspondence between$\left [ 0,1 \right ]\bigcup \left [ 2,3 \right ]\bigcup ..$ and $\left [ 0,1 \right ]$ 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 
 A: 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. 
A: 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:
$$0\overset{f}\mapsto\frac12\overset{f}\mapsto\frac14\overset{f}\mapsto\frac18\overset{f}\mapsto\frac1{16}\overset{f}\mapsto\dots\;.$$
