Proving that (0,1) and [0,1] are numerically equivalent. as the title suggests, I need help proving that the cardinality of $(0,1)$ and $[0,1]$ are the same. 
Here is my work: 
$f:[0,1] \rightarrow (0,1)$
Let $n\in N$
Let $A=\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}....\}\cup \{0\}$
On $[0,1]\in A: f(x)=x$
On $A: f(0)=\frac{1}{3}$
$f(1)=\frac{1}{2}$
$f(\frac{1}{n})=\frac{1}{n+2}, n>2$
Now I will prove that $f$ is a bijective function.
Let $f(n)=f(m)$ for $n,m \in N$ and $n,m>2$. Then $\frac{1}{n+2}=\frac{1}{m+2}$. We multiply both sides of the equation by $(n+2)(m+2)$ and obtain $m+2=n+2 \rightarrow m=n$. Thus $f$ is injective.
From here on out, I am kind of shakey. I know the gist of this proof, but I don't know how to set up the sequencing correctly. For instance, I know that I need to map $x_1$ to $0$ and $x_2$ to $1$ and $x_{n}$ to $x_{n+2}$, but I am confused on how to do so. 
 A: Your proof has some problems.


*

*You haven't defined $A$. I suppose it should denote some subset of $[0,1]$.

*If $A$ is a subset of $[0,1]$ then it's not the case that $[0,1]\in A$. Perhaps you meant $[0,1]\setminus A$?

*The idea is that $A$ can be "easily described" as a sequence $a_n$ for $n\in\Bbb N$, preferably such that $a_0=0$ and $a_1=1$. Then you can define a function which maps $a_n\mapsto a_{n+2}$, and $x$ for those not in $A$ (as you did).


Then to show that the map is a bijection you need to show that if $x,y\in[0,1]$ are distinct then $f(x)\neq f(y)$. This is done by dividing into cases, both $x,y\in A$, both $x,y\notin A$ and the case that $x\in A$ and $y\notin A$ (there's a fourth case which is similar to this one).
Then you have to show that every $y\in(0,1)$ is in the range, which is also done by dividing into cases, if $y\notin A$ then $f(y)=y$; and if $y\in A$, then you have to do some work.
A: I think you're getting a little too bogged down in the details. The basic idea is that the sets $$A_0=\{a_0,a_1,a_2\ldots,a_n,\ldots\}\text{ and }A_1=\{a_1,a_2,a_3\ldots,a_n,\ldots\}$$ have the same cardinality, where $\{a_0,a_1,a_2\ldots\}$ is any sequence of distinct numbers. So $$B\cup A_0\text{ and }B\cup A_1$$ have the same cardinality, where $B$ is any set disjoint from $A_0$ (and hence also disjoint from $A_1$). This shows that adding one element to the set $B\cup A_1$ doesn't change the cardinality.
So if you start with some infinite set $S$ and some element $a_0$ not in the set, the trick is to express $S$ as $B\cup A_1$. Well, just pick an infinite sequence in $S$ to serve as $A_1$, and let $B=S-A_1$.
The sets $(0,1)$ is an example of such an infinite set $S$, so it follows that $[0,1)$ has the same cardinality as $(0,1)$. Repeat to get $[0,1]$ with the same cardinality. Or modify the previous argument to show that adding two elements doesn't change the cardinality of an infinite set: $$A_0=\{a_0,a_1,a_2\ldots,a_n,\ldots\}\text{ and }A_2=\{a_2,a_3,a_4\ldots,a_n,\ldots\}$$ have the same cardinality, etc.
Now your proof follows this idea; you've just made a particular choice for the sequence $\{a_2,a_3,\ldots\}$, namely $\{\tfrac{1}{2}, \tfrac{1}{3},\ldots\}$. Also you are trying to construction the one-one correspondence explicitly.
This isn't hard. We want $f:B\cup A_0\rightarrow B\cup A_2$ one-one and onto. To begin with, as you indicated, we want $f(a_0)=a_2$, $f(a_1)=a_3$, etc.; i.e., $f(0)=\tfrac{1}{2}$, $f(1)=\tfrac{1}{3}$, and $f(\tfrac{1}{n})=\tfrac{1}{n+2}$ for $n\geq 2$. Then for $x\not\in \{a_0,a_1,\ldots\}$, let $f(x)=x$.
(I'm assuming when you wrote "On $[0,1]\in A$", that was a typo for "On $[0,1]-A$". Also, I'm using $-$ for set-difference, what some people denote by $\smallsetminus$.)
