Show the continuous image of a compact set is compact.

Question

Using the open-cover definition of compactness (if X is a topological space, a collection of sets {$U_{\alpha}|\alpha\in A$} with each being open in X, is said to be an open cover of X if $X=\bigcup_{\alpha\in A}U_\alpha$. The space $X$ is compact if for every open cover of $X$, there is a finite subcollection of {$U_\alpha$} that is also an open cover of X.) prove that the continuous image of a compact set is compact. This implies compactness is a topological property? The exercise hints at starting with an arbitrary open cover of the image space. Any further hints would be appreciated. Thanks in advance!

Edit: I believe the possible duplicate question concerns a less general case in $R^n$. Also, the answers do not address open covers which is the source of much of my confusion.

Answer 1

So suppose we have a compact space $X$ and a continuous surjective $f:X \to Y$, and we want to show $Y$ is compact:

Let $\mathcal{U} = \{O_a, a \in A\}$ be an arbitrary open cover of $Y$, indexed by some set $A$.

Then define $\mathcal{V}:=\{f^{-1}[O_a]: a \in A\}$.

All members of $\mathcal{V}$ are open subsets of $X$, as $f$ is continuous and so inverse images of open sets in $Y$ are open in $X$.

It's also a cover of $X$: if $x \in X$, then $f(x) \in Y$ and so there is some $b \in A$ such that $f(x) \in O_b$, as $\mathcal{U}$ is a cover of $Y$. But then $x \in f^{-1}[O_b] \in \mathcal{V}$.

As $X$ is compact, there are finitely many $a_1, \ldots, a_N \in A$ such that $\mathcal{V}':= \{f^{-1}[O_{a_i}]: i= 1,2, \ldots, N\}$ is also a cover of $X$ (every open cover of $X$ has a finite subcover).

Then $\mathcal{U'} = \{O_{a_1}, \ldots, O_{a_N}\}$ is a finite subcover of $\mathcal{U}$: suppose $y \in Y$. As $f$ is surjective we have some $x \in X$ such that $f(x) = y$ and as $\mathcal{V}'$ is a cover, there is some $a_i, 1\le i \le N$ such that $x \in f^{-1}[O_{a_i}]$, which means that $y = f(x) \in O_{a_i} \in \mathcal{U}'$, so that $y$ is indeed covered. So we are done and $Y$ is compact.

Answer 2

$X=f^{-1}\left(\displaystyle\bigcup_{i}G_{i}\right)=\displaystyle\bigcup_{i}f^{-1}(G_{i})=f^{-1}(G_{i_{1}})\cup\cdots\cup f^{-1}(G_{i_{N}})$, then $Y=f(X)\subseteq G_{i_{1}}\cup\cdots\cup G_{i_{N}}$.