Find a bijection between $[1,2)$ and $(1,2)$ I want to find a bijection between $[1,2)$ and $(1,2)$ and prove it.
My attempt:
$[1,2) = \{x \in \mathbb R | 1 \leq x <2\}$
$(1,2) = \{x \in \mathbb R | 1 < x < 2\}$
$f(x) = x$ if  $x \ne 1\frac{1}{n}$ for $n = 1,2,3,...$
and
$f(x) = 1\frac{1}{x+1}$, if $x =1\frac{1}{n}$ for $n = 1,2,3,...$
Proof - Injective - Prove $x_1 = x_2$ for $f(x) = x$
\begin{align*}
  f(x_1) = f(x_2) &\implies x_1 = x_2.\\
\end{align*}
Proof - Injective - Prove $x_1 = x_2$ for $f(x) = 1\frac{1}{x+1}$
\begin{align*}
  f(x_1) = f(x_2) &\implies 1\frac{1}{x_1+1} = 1\frac{1}{x_2+1}\\
 &\implies \frac{1}{x_1+1} = \frac{1}{x_2+1}\\
 &\implies x_2+1 = x_1+1\\
 &\implies x_2 = x_1.\\
\end{align*}
Therefore $f$ is injective.
Any help will be appreciated.
Thanks.
 A: This is the right idea, but the details are not quite right. For example, you say
$$f(x) = 1\frac{1}{x+1}\text{ if }x =1\frac{1}{n}\text{ for }n = 1,2,3,...$$
I think you meant to say $f(x) = 1\frac1{n+1}$ there on the left.  I will suppose that you meant that.  if you really meant $f(x) = 1\frac1{x+1}$, that is an additional problem.
Supposing that you meant $f(x)=1\frac1{n+1}$, then when $n=1$ you've said that $f(2)=1\frac12=\frac32$. But you shouldn't be defining $f(2)$ at all, because $2$ is not an element of $[1,2)$. At the other end of the interval, $1$ does not have the form $1\frac1n$, so you've defined $f(1)=1$.  But $1$ is not in the desired range $(1,2)$.
Also it's not enough to prove that $f$ is injective; you must also prove that $f$ is surjective: for each $y$ in $(1,2)$, there is some $x$ in $[1,2)$ for which $f(x)=y$.
Check the details a little more carefully.
Also you should know that nobody writes $1\frac1n$ to mean $1+\frac1n$.
A: In your question, you noted that card((1,2)) <= card([1,2))
Let $f: [1,2)\to(1,2)$
Fix $x \in (1,2)$. Let $\{A_i\}_{1}^{\infty}$ represent the non-terminating decimal expansion right of the decimal place (this is unique to each number).
Let $f(x) := 1 + \sum_{i=1}^{\infty}A_i*10^{-i-1}$. All values of f are of the form 1.0... Therefore, let $f(1):=1.1$. As each decimal expansion is unique the resulting function f is injective.
Therefore, there exists a bijection between (1,2) and [1,2)
