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Let $X$ and $Y$ be topological spaces and $f:X\to Y$. Prove that $f$ is continuous if and only if $\overline{f^{-1}(B)}\subset f^{-1}(\overline B),$ for each $B\subset Y$.

  • $\rightarrow$

Let $B\subset Y$. Then

$B\subset \overline B$

$\implies f^{-1}(B)\subset f^{-1}(\overline B)$

$\implies\overline{ f^{-1}(B)}\subset\overline {f^{-1}(\overline B)}=f^{-1}(\overline B),$ by the continuity of $f$.

  • $\leftarrow$

Let $\overline B\subset Y,$ we want to show that $f^{-1}(\overline B)$ is closed, i.e. $\overline{ f^{-1}(\overline B)}\subset f^{-1}(\overline B).$

$$f^{-1}(B) \subset\overline{f^{-1}(B)}\subset f^{-1}(\overline B) $$

I am stuck here, what can I do from here?

Note that it was just an arbitrary idea to show continuity using closed sets, but if there is an easier way to show continuity, please suggest it.

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  • $\begingroup$ ah how come I didn't see $B=\overline B$.. $\endgroup$ – elli saba Feb 23 '18 at 0:44
  • $\begingroup$ You can try using sequences. Pick up a sequence $a_n \to a\in X$ and show that $f(a_n)\to f(a)\in Y$. $\endgroup$ – Valerin Feb 23 '18 at 0:49
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For the converse, assume that for every $B \subset Y$, we have $\overline{f^{-1}(B)} \subset f^{-1}(\overline{B})$. Now pick any $K$ closed in $Y$. To show $f^{-1}(K)$ is closed in $X$. This will show continuity of $f$. The relation $f^{-1}(K) \subset \overline{f^{-1}(K)}$ always holds. So we must show that $\overline{f^{-1}(K)} \subset f^{-1}(K)$. By our assumption, we have that $\overline{f^{-1}(K)} \subset f^{-1}(\overline{K}) = f^{-1}(K)$ (since $K$ is closed in $Y \Rightarrow \overline{K} = K$).

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    $\begingroup$ so obvious.. I didn't see it :( $\endgroup$ – elli saba Feb 23 '18 at 0:46
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    $\begingroup$ Happens to all of us :) hope this helps. $\endgroup$ – Fabian Feb 23 '18 at 0:47

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