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?

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    $\begingroup$ How do you prove that the image of a compact (just compact) set is compact? Imitate. $\endgroup$ – User May 23 '18 at 20:53
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    $\begingroup$ The image of an open set under a continuous map need not be open, but the inverse images of open sets are open. This is key $\endgroup$ – Shaun_the_Post May 23 '18 at 20:55

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.

  • $\begingroup$ Thank you! Everything is clear $\endgroup$ – user557550 May 24 '18 at 18:29
  • $\begingroup$ @MariyaKav Glad to help! Maybe you could mark the answer as accepted? $\endgroup$ – Henno Brandsma May 24 '18 at 18:30
  • $\begingroup$ Sure, i did!.., $\endgroup$ – user557550 May 27 '18 at 13:56

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