Understanding the proof of "continuous image of compact set is compact"

Question

I have a trouble understanding the proof of theorem.

Theorem. If a function $f:K \rightarrow \mathbb{R}$ is continuous and $K$ is compact set, then $f(K)$ is compact too.

Proof. (This is part of proof in my text)

Let $\{V_\alpha \mid \alpha \in A\}$ is an open cover of $f(K)$, then $\{f^{-1}(V_\alpha) \mid \alpha \in A \}$ is also open cover of $K$.

I agree that it works because of the theorem that for continuous function and $\forall G \subseteq \mathbb{R}$ which is open set, $f^{-1}(G)$ is also open set on domain. Moreover if we denote $C = \cup V_\alpha$, then $f(K) \subseteq C$, $f^{-1}(f(K)) \subseteq f^{-1}(C)$. It is also known that $K \subseteq f^{-1}(f(K))$ then $K\subseteq f^{-1}(C)=f^{-1}(\cup V_\alpha) = \cup f^{-1}(V_\alpha)$. Then the statement in the proof is true that $\{ f^{-1}(V_\alpha) \mid \alpha \in A \}$ is also open cover of $K$.

Also, for each $\alpha$, there exists open set $U_\alpha$ such that $U_\alpha \cap K = f^{-1}(V_\alpha)$.

I think it works by the property of relatively open set that $f^{-1}(V_\alpha)$ is open in $K$, so there exists $U_\alpha \cap K = f^{-1}(V_\alpha)$.

Then $\{U_\alpha \mid \alpha \in A\}$ is open cover of $K$.

How can we verify that the last sentence is true? Intuitively I agree with it but I can't logically explain why it makes sense.

Answer 1

If $x\in K$ then $f(x)\in V_{\alpha_x}$ for some $\alpha_x\in A$. This because the $(V_{\alpha})_{\alpha\in A}$ cover $f(K)$.

Then $x\in f^{-1}(V_{\alpha_x})$.

Evidently for any $x\in K$ we can find an $\alpha_x\in A$ such that $x\in f^{-1}(V_{\alpha_x})$, so the $(f^{-1}(V_{\alpha}))_{\alpha\in A}$ cover $K$.

Answer 2

You might be getting confused by the relative topology. There is no relative topology here as $K$ is the only space in our initial space.

The continuity of $f$ is equivalent to the preimage of any open set $V \subset \mathbb{R}$ being open in $K$. In particular consider any open cover of $f(K)$. The open cover will be collection of open sets in $\mathbb{R}$. If we denote the union of this collection as $V$ then clearly we have $$ f(K) \subset V$$ Furthermore preimages preserve the subset so we have $$ K \subset f^{-1} V$$ Now note that the preimage of a collection of sets is equal to a collection of preimages of the sets so we have $$ f^{-1} V = \bigcup_{\alpha \in A} f^{-1} V_\alpha$$ So we have $$ K \subset \bigcup_{\alpha \in A} f^{-1} V_\alpha$$ By continuity of $f$ this is a collection of open sets which covers $K$. By compactness of $K$ we know that there exists some finite covering of $K$. Let $B$ be a finite set by the above we now have $$ K \subset \bigcup_{\beta \in B} f^{-1} V_\beta \subset \bigcup_{\alpha \in A} f^{-1} V_\alpha$$ Again note that $$ \bigcup_{\beta \in B} f^{-1} V_\beta = f^{-1} \bigcup_{\beta \in B} V_\beta \subset f^{-1} \bigcup_{\alpha \in A} V_\alpha$$ So we have $$ K \subset f^{-1} \bigcup_{\beta \in B} V_\beta $$ Taking the image of both sides preserves the subset so we have $$ f(K) \subset \bigcup_{\beta \in B} V_\beta \subset \bigcup_{\alpha \in A} V_\alpha$$ Since the open cover $V$ was arbitrary we have shown that there exists a finite open subcover for each $V$ and hence $f(K)$ is compact.