No Bijection from set $X$ to $X - \{x\}$ I want to show that there is no bijection from finite set $X$ to the set $X - \{x\}$. But I don't want to use any of the rules of cardinality or Cantor–Bernstein–Schroeder theorem etc.
Is there a more basic proof ?
 A: Well you can prove this by induction on the number of elements in $X$.
If $X$ has one element, then this is obvious, because then $X=\{x\}$ and $X\setminus\{x\}=\varnothing$. Clearly there is no injection from a non-empty set into the empty set.
Suppose this holds for sets of $n$ elements, i.e. if there is an injection from a set of $n$ element into itself then it is a surjection. Let $X$ be a set of $n+1$ elements, and pick some $x\in X$.
Let $X'=X\setminus\{x\}$, then $X'$ is a set with $n$ elements. If there was an injection $f\colon X\to X'$ then the restriction of $f$ to $X'$ (denoted by $f\upharpoonright X'$) is an injection from $X'$ into itself, so by our induction hypothesis it has to be surjective. Let $x'=f(x)$ then $x'\in X'$.
Because $f\upharpoonright X'$ is surjective there is some $y\in X'$ such that $f(y)=x'$, but now we have that $f(y)=f(x)$ and since $f$ was injective we have that $y=x$ which is a contradiction because $x\notin X'$. Therefore there is no such injection.
A: This one is wrong, cause for example there are bijections from $[0,1]$ to $(0,1]$.
For finite sets it is true, but I don't see how a proof shall work without cardinality arguments. I hope you will correct me if i am wrong, but the only difference between a finite and an infinite set is its cardinality, so if you don't use this one, your proof would imply that for every set this is true. But as i mentioned, for infinite sets it isn't true so you must use the cardinality. 
Edit: Maybe there is a proof without cardinality, as I didn't listen to topology I won't be able to do it rigorous, but let's take this Definition of finite:
We call a set $A$ finite if the topological space $(A,T)$ is Hausdorff iff $T$ is the discrete topology.
Any bijection between discrete finite Topological spaces is a homoeomorphism. Now maybe we can show that one of the homotopie groups aren't equal (since the 0th is just a set we have to take a higher one).
If I am right we have to show that the free group with $n$ and $n-1$ generators aren't the same, maybe this works without cardinality. 
