Let $(X,\tau),(Y,\eta)$ be topological spaces. If $f:X\to Y$ is continuous, $f(X)=Y$ and $X$ is compact, then $Y$ is compact.
We want to show that for every open cover for $Y$ there exist a finite open subcover. So let $\{U_\alpha:\alpha\in I\}$ be ab open cover for $Y$. Thus $Y=\bigcup_{\alpha\in I} U_\alpha$
Now how can I use the hypothesis that $f$ is continuous?
I was thinking about this property of continuity: $\forall U\in\eta,f^{-1}(U)\in\tau$ but I don't know how to make a relation with the compactness of $X$.
As $X$ is compact, then $\forall U\subset\tau,X\subset\bigcup U$ and $\exists u_1,…,u_n\in U:U\subset\bigcup_iU_i$.
Could anyone help me please?
Best Answer
Consider the collection of sets $f^{-1}(U_{\alpha})$ for all $\alpha$.
They are open as $f$ is continuous and cover $X$ since $f$ is surjective.
Since $X$ is compact take a finite subcover $f^{-1}(U_{\alpha_1}), \ldots, f^{-1}(U_{\alpha_n})$.
What can you now deduce about $U_{\alpha_1}, \ldots, U_{\alpha_n}$?