Find a bijective mapping that shows that [0,1] and [0,1) have the same cardinality I need to show that the two sets $[0,1]$ and $[0,1)$ have the same cardinality. I know that in order to show this I must show that there exists $f$ such that $f:[0,1]\to[0,1),$
but I am not sure how to proceed.
Any help would be appreciated. Thanks.
 A: These sorts of things often boil down to using the fact that there is a bijection between the nonnegative integers and the positive integers, allowing you to shift everything over by one to make room for an extra point (e.g. the Hilbert hotel paradox); you just need to find a suitable copy of $\mathbb{N}$ in the problem.
In fact, we can completely characterize the bijections:
Lemma: every bijection between $[0,1]$ and $[0,1)$ can be uniquely expressed as:


*

*An injection $g : \mathbb{N} \to [0,1)$

*A bijection $h : [0,1) \setminus g(\mathbb{N}) \to [0,1) \setminus g(\mathbb{N})$


and conversely, given such data, there there is a corresponding bijection
$$ f : [0,1] \to [0,1) : x \mapsto
\begin{cases}
g(0) & x = 1 
\\ g(g^{-1}(x) + 1) & x \in g(\mathbb{N})
\\ h(x) & x \in [0,1) \setminus g(\mathbb{N})\end{cases} $$
A: Suppose $A \subset [0,1]$ is countably infinite and $1 \in A$. Then $[0,1] \setminus A=[0,1) \setminus A$. So the identity function, call it $f$, is a bijection from $[0,1] \setminus A$ to $[0,1) \setminus A$. Can you define a bijection, call it $g$, from $A$ to $[0,1) \cap A$? Once you have done that,
$$h(x)=\begin{cases} g(x) & x \in A \\
f(x) & x \not \in A \end{cases}$$
is a bijection from $[0,1]$ to $[0,1)$. A hint for constructing $g$: consider enumerating $A$ as a sequence $\{ x_n \}_{n=1}^\infty$ with $x_1=1$.
A: Define:
\begin{equation} f(x)=\begin{cases} \frac{1}{1+n}, & \text{if $x = \frac{1}{n}$ , $n \in \mathbb{N}$ }.\\ x , & \text{otherwise}. \end{cases} \end{equation}
