Bijection between open and closed interval

Question

I am not sure how to approach the following problem:

Show the open interval $(a,b)$ is bijective with the closed interval $[c,d]$.

I was thinking of using $a+u$ where $u$ is a really small number and $b+v$ where $v$ is a really small number to make a new interval $[a+u, b-v]$, which is equivalent to (a,b).

Answer 1

List all rationals on $(a,b)$ say, $\{r_n:n=1,2,\dotsc\}$. Let $r_{-1}=a$ and $r_0=b$. Fist define a map $\varphi:(a,b)\to [a,b]$ as follows:

$\varphi(x)= x$, if $x$ is irrational and $\varphi(r_n)=r_{n-2}$, for each $n=1,2,\dotsc$. Clearly $\varphi$ is a bijection.

Also define a map $\psi:[a,b]\to [c,d]$ by $\psi(x)=\left(\dfrac{d-c}{b-a}\right)x+\left(\dfrac{bc-ad}{b-a}\right)$. Then also $\psi$ is a bijection. Clearly the composition map $f=\psi\circ\varphi:(a,b)\to [c,d]$ is the required bijection.

Answer 2

You can use Cantor-Schröder-Bernstein theorem, which states if you can find an injective function from (a,b) to [c,d] and vice versa, there exists a bijection between them.

http://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

Answer 3

It is enough to find a bijection $f:[0,1] \to (0,1)$. $$f(x) = \left\{ \begin{array}{1 1} 1/2 & \mbox{if } x =0 \\ 1/2^{n+2} & \mbox{if } x = 1/2^n\\ x & \mbox{otherwise} \end{array} \right.$$ does the job.