Is it possible to prove that the cardinality of a set is unique using only the definition of bijection and cardinality. By definition, the function $f$ from $X$ to $Y$ is a bijection if every element $y$ in $Y$ is a value of $f(x)$ for exactly one $x$ in $X$. Also by definition, if there is a bijective correspondence between the sets $S$ and $\mathbb{N_m}$ then S has cardinality $m$, which can be denoted by $|S|=m$.
I want to prove that if we have a set S such that $|S|=s$ and $|S|=t$, then $s=t$.
Given is that there are bijections $f:S\to\mathbb{N_s}$ and $g:S\to\mathbb{N_t}$. Now we can use the pigeonhole principle for the composite functions $f\circ g^{-1}$ and $f^{-1}\circ g$ to obtain $s\le t$ and $t\ge s$, respectively.
Now my question is: why can we not say directly, from the definition of a bijection, that the cardinality of $X$ and $Y$ are equal. So that we can prove it by saying that if $|S|=s$ and $|S|=t$ then $s=t$ because both $f$ and $g$ are bijections with the same domain.
 A: What you want to prove is that if $\mathbb N_s\cong\mathbb N_t$ then $s=t$. WLOG we can assume that $s\le t$.
Basically you may prove this by showing that there can be no smallest such cardinality that allows this, because if you find one you can create a case with one element less - unless one is $\emptyset$. You end up with the claim that $\emptyset\cong X$ for some non-empty $X$.
Formally one starts with showing that cardinality is a total order. So if $X\prec Y$ and $Y\preceq Z$ we have that $X\prec Z$. We can use this to show that if $\mathbb N_s \prec \mathbb N_{s+1}$ then it's true for all $t>s$ that $\mathbb N_s\prec\mathbb N_t$. That is we only have to show that $\mathbb N_s \prec\mathbb N_{s+1}$.
Next is to use induction. Obviously $\emptyset=\mathbb N_0\prec\mathbb N_1$ (since to have a bijection with the empty set the other need to be empty). Now assume it's true for some $s$. Now assume that we have $\mathbb N_{s+1}\cong\mathbb N_{s+2}$ that is there's a bijettion $\Phi$, but then we can restrict that to $\phi:\mathbb N_s\to\Phi(\mathbb N_s) = \mathbb N_{s+2}\setminus \{\Phi(s+1)\}$. Now we use this latest form to form a bijection to $\mathbb N_{s+1}$ which would show that $\mathbb N_s\cong\mathbb N_{s+1}$ which we assumed not to be possible. 
