# Proof Verification: Using Heine-Borel property to show continuous function maps compact set to compact set.

Let $$f : M \mapsto N$$ be continuous. Consider a compact $$K \subseteq M$$.

Assume as given: $$\forall B \subseteq N$$ open, $$f^{-1}(B)$$ is open in $$M$$.

1. Consider an open cover of $$f(K)$$, call it $$A_i$$ for $$i \in I$$.
2. Then, $$B_i$$ is an open cover of $$K$$, where $$B_i = \{x \in M: f(x) \in A_i \}$$
3. There exists a finite subcover of $$K$$, call it $$B_{n_1}, ..., B_{n_k}$$
4. Then, $$A_{n_1}, ..., A_{n_k}$$ is a finite subcover of $$f(K)$$

Just need to verify this proof, in particular, I am concerned if $$(2)$$ and $$(4)$$ need more details.

1. What do you mean by calling the open cover $$A_i$$? I would start 'Let $$U$$ be an open cover of $$f(K)$$.'
2. Yes, more details would be welcome. Why is what you denote $$B_i$$ an open cover of $$K$$? Answer: for every $$x \in K$$, $$f(x)$$ belongs to some $$u \in U$$.
4. Again, more details would be helpful. Answer: for every $$y \in f(K)$$ we have $$y = f(x)$$ where $$x \in K$$. Then $$x \in f^{-1}(U_i)$$ for some $$i$$ and so $$y = f(x) \in U_i$$.
(2) works out because $$B_i = f^{-1}(A_i)$$ is open for all $$i$$ since $$f$$ is continuous, and certainly covers $$K$$, so it is indeed an open cover as you stated.
In a similar fashion for statement (4), $$A_{n_1}, \dots, A_{n_k}$$ is a finite sub cover since for any $$y \in f(K), \; \exists i : x \in B_{n_i}$$ (Since $$\{B\}_{i \in I}$$ is an open cover for $$K$$) so that $$f(x) =y \in A_{n_i}$$ by construction.