I want to show that if $B\subset A$ and $f:A\to B$ is an injective function then there's a bijection between $A$ and $B$.
I believe my "proof" is wrong, I probably use too much "intuition" when I try to solve it. But hopefully I will get a better feeling if someone tells me where/what I do wrong and help me. 🙂 That said, a friend to me "solved" another problem
"if $f:A\to C$ is an injective function, and $g:C\to A$ is an injective function, then there is a bijection between $A$ and $C$".
She argued like this " for all $a$ in $A$ we can find an element $f(a)$ in $C$ and for all $c$ in $C$ we have an element $g(c)$ in $A$. If f(a)=c then we must have g(c)=a. This holds for all $a$ and $c$. So each $a$ maps to exactly one $c$ and each $c$ maps to exactly one $a$".
That is the Schröder–Bernstein theorem though, I have seen the proof, so I could directly tell that it's a wrong proof. I would probably not argue in the exactly same way but probably in a similar fashion. On the other hand I cannot really tell why this doesn't prove the fact either. Oh well, here comes my proof, it's a similar argument so I guess I'm wrong 🙂
Proof:
Since I already know it's injective, I just have to show it's surjective. We have that $B\subset A$, that is, every element of $B$ is in $A$. Because of this we can for every element $b\in B$ find an element $a\in A$ such that $b=f(a)$. That is, $\forall b\in B\exists a\in A:\textbf{ }b=f(a)$. But that is the definition of surjection. Hence, there exists a bijection between $A$ and $B$ since f is injective.
I bet I've forgot to mention something now, which I found important to mention, but unfortunately I have forgotten it. Hopefully I will remind myself. Thanks for your help 🙂
Best Answer
It doesn't hold that $f$ gives the bijection. Consider $A=(0,2),B=(0,1)$ and $f(x)=\dfrac{x}{3}.$
The problem in your proof: Each element of $B$ is an element of $A,$ but this doesn't mean that an element of $B$ has a preimage. Think of the above example.