Closed Immersions are open in the strong topology $C_S^r(M,N)$

Question

I am trying to show that closed immersions are open in the strong topology $C_S^r(M,N)$. I know that a proper map $f$ into a locally compact haussdorf space, which $N$ is , will be closed . So now I am just trying to see that if I have a closed immersion between manifolds then it will be a proper map. If I have this then I have that the set of closed Immersions is just the intersection of the open set of Immersions with the one of proper maps and so it is open.

So does anyone know why a closed immersion is a proper map ? I have tried seeing this using subsequences but I got nowhere, so any hints are aprecciated.

Thanks in advance.

Answer 1

After thinking about this for a while I think I have come up with a solution:

So let's suppose that we have a closed immersion $f: M\rightarrow N$, that is not a proper map, where $M$ and $N$ are manifolds. So there exists a compact set $K\subset N$ such that $f^{-1}(K)$ is not compact,and so there exists a sequence $\{x_n\}_{n=1}^{\infty}$ that has no convergent subsequences. Now since $f$ is an immersion there will exists neighborhoods of $x_n$ such that $f|_{V_n}$ is injective. Since $K$ is compact there exists a subsequence $f(x_{n_k})\rightarrow y\in N$. Now using the fact $f|_{V_n}$ is injective we can choose $z_k$ such that $f(z_k)\neq y$, $d(z_k,x_{n_k})<\frac{1}{k}$ and $f(z_k)\rightarrow y$. Now since we choose $\{z_k\}$ in a way that $d(z_k,x_{n_k})<\frac{1}{k}$ we have that $B=\{z_k\}_{k=1}^{\infty}$ is a closed set, since it has no convergent subsequences. Also note that $y\in \bar f(B)-f(B)\neq \emptyset$ and so we obtain a contradiction with the fact that $f$ is a closed map, hence proving the dersired result.