Let A,B be finite sets. Assume size(A)=size(B) and A$\subseteq$B then A=B.
Proof: Since size(A)=size(B), then there exists a one to one correspondence between the two sets. Assume that A$\neq$B, that means that $\exists$ b$\in$B such that f(a)$\neq$b that means that the function is not surjective, which is a contradiction.
Is this proof correct? Furthermore, is there another way to prove it? Am I correct to say that the same proof works for countable sets?
Best Answer
You don't need a one-to-one correspondence, you can just do the following.
Suppose $|A|=|B|$ and $A\subseteq B$. Assume that $A\neq B$. Then there exists $b\in B$ such that $b\not\in A$. But every element of $A$ is in $B$, which implies that $|A|\leq |B|-1$, so $|A|<|B|$, a contradiction.