I want to prove the following
Let $(M,d_M)$ and $(N,d_N)$ be some metric spaces. If $M$ is compact then a continuous bijection $f:M\to N$ is a homeomorphism. In other words, its inverse $f^{-1}:N\to M$ is automatically continuous.
I want to write a proof by using the sequential definition of compactness and avoiding arguments by contradiction (Is it possible? I don't know!). I already know a proof for this which uses contradiction.
My Attempt. Let $(q_n)$ be a convergent sequence in $N$ so $q_n\to q$ with $q\in N$. To show that $f^{-1}:N\to M$ is continuous we would like to verify that $f^{-1}(q_n)\to f^{-1}(q)$. Since $f$ is a bijection there are unique $(p_n)$ and $p$ such that $f(p_n)=q_n$ and $f(p)=q$. So our aim is to show that $p_n\to p$.
$(p_n)$ is a sequence in $M$ and since $M$ is sequentially compact then every sequence in it should have a convergent sub-sequence in $M$. Consequently, there is a convergent sub-sequence $p_{\phi_n}\to p_*$ with $p_*\in M$. Since $f$ is continuous we have $f(p_{\phi_n})\to f(p_*)$, which is equivalent to $q_{\phi_n}\to f(p_*)$. Knowing that any sub-sequence of a convergent sequence converges to the same limit of the mother sequence, we get $f(p_*)=q=f(p)$. Since $f$ is injective, this implies that $p=p_*$. So far, we have proved that $p_{\phi_n}\to p$.
This is where I got stuck. I don't have any idea to proceed further. Any hint or help is appreciated.
Best Answer
I think your argument can be easily modified using the following fact:
A sequence $(x_n)$ in $X$ converges to some $x\in X$ if and only if every subsequence of $(x_n)$ has a further subsequence that converges to $x$.
So you can start with an arbitrary subsequence of $(p_n)$ and conclude that every subsequence of $(p_n)$ has a further subsequence which converges to $p$. Then according to the fact, $(p_n)$ converges to $p$.