Sequential compactness and continuity

Question

Let $K\subset \mathbb{R}^n$ be sequentially compact and let $F:K\to \mathbb{R}^m$ be continuous and onto. Prove that $A\subset\mathbb{R}^m$ is closed (in $\mathbb{R}^m$) if and only if $F^{-1}(A)\subset K$ is closed (in $\mathbb{R}^n$). Then prove that $F^{-1}:\mathbb{R}^m\to\mathbb{R}^n$ is continuous.

$\pmb{A\subset\mathbb{R}^m}$ is closed

$\Rightarrow$

Suppose that $A$ is closed in $\mathbb{R}^m$.

Let $\{u_k\}$ be a sequence of points in $F^{-1}(A)$. Then the points are in $K$.

Since $A$ is closed, every sequence in it converges to a point in $A$.

Since $A$ is a subset of $\mathbb{R}^m$, every point in it was mapped there by at least one point in $K$.

I want to use the fact that since $F$ is continuous, $F(K) = \mathbb{R}^m$ is sequentially compact, so.... maybe that can help? I am going in circles on this one. I can't write anything coherent about it.

Answer 1

The forward implication is straightforward from continuity. $A \subset \mathbb R^m$ is closed in $R^m$, and $f:\mathbb R^n\to \mathbb R^m$ is continuous,then $f^{-1}(A)$ is closed in $\mathbb R^n$. Thus closed in $K$.

$\Leftarrow$

Suppose $f^{-1}(A)$ is closed in $K$. Let $\{a_n\}$ be a sequence in $A$.

Then choose $b_n=f^{-1}(a_n).$ Since $K$ is sequentially compact and $f^{-1}(A)$ is closed in $K$, $f^{-1}(A)$ is sequentially compact.

Since $b_n\in f^{-1}(A)$ there exist subsequence $\{b_{n_k}\}_{k\in \mathbb Z^+}$ of $\{b_n\}$ s.t $\lim\limits_{k\to\infty}b_{n_k}= b\in f^{-1}(A)$.

Thus $f(b_{n_k})$ is a subsequence of $\{a_n\}$ s.t $f(b_{n_k})\to f(b)$ as $k\to \infty.$

$(\lim\limits_{k\to\infty}f(b_{n_k})= f(b)\ \text{form continuity of}\ f)$