Let $X$ be a space that is countably compact, which means that every countable open cover has a finite subcover. Assume that $f$ is a continious function. Then $f(A)$ is contably compact.
So, take a space that is countably compact. Denote $A := f(X)$. Take a countable open cover of A. I have to show now that it has finite subcover. But don't have any idea on how to proceed further. Does it follow from the fact that image of open sets are open if $f$ is continious. I mean since we know that $f$ is continious function then image of finite open subcover of X will be open as well in A. Or it is not enough?
Could anyone please give some hints?
This proof is just a straightforward proof-idea copy of the compact case:
Let $\{U_n: n \in \omega\}$ be a countable open cover of $f[X]$ by open subsets of $Y$, assuming $f:X \to Y$ is our continuous function, for notation's sake.
Then $\{f^{-1}[U_n]: n \in \omega\}$ is a countable open cover of $X$ (the sets are open by continuity of $f$; if $x \in X$ some $U_m$ must cover $f(x) \in f[X]$, and then by definition $x \in f^{-1}[U_m]$, so all $x$ are covered). Finitely many of these cover $X$, as $X$ is countably compact, say $f^{-1}[U_{n_1}],\ldots, f^{-1}[U_{n_m}]$ do.
But then $f[X]$ is also covered by the $U_{n_1}, \ldots, U_{n_m}$ (any $y \in f[X]$ is of the form $y =f(x)$ for some $x \in X$, $x$ lies in some $f^{-1}[U_{n_j}]$ from the finite subcollection, and then $y = f(x) \in U_{n_j}$ by definition, so is covered).
So every countable open cover of $f[X]$ has a finite subcover.