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Let $X$ and $Y$ be topological spaces. Suppose that $f: X \rightarrow Y$ is a function satisfying: For every $A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$.
Prove $f$ is continuous.

$\textbf{Proof:}$ Let $X$ and $Y$ be topological spaces. Suppose $f: X \rightarrow Y$ is a function satisfying $\forall A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$.
Let $a \in \overline{A}$ and then $f(a) \in \overline{f(a)}$.
Let $V \subset Y$ be open with $f(a) \in V$.
By way of contradiction, suppose $\forall \, \text{open} \, U \subset X$ with $a \in U$, $f(U) \not \subset V$.
So, $\forall a \in U$, $b \in X \backslash f^{-1}(V)$ is also $b \in U$.
Then, $a \in \overline{X \backslash f^{-1}(V)} \subset \overline{Y \backslash V}$ $\rightarrow \leftarrow$ (contradiction!) because $V \cap Y \backslash V = \emptyset$.
$\therefore f$ is continuous. $\,\, \blacksquare$

symplectomorphic
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Norman Contreras
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    An easier, more direct way is this: https://math.stackexchange.com/a/114572/732532 – morrowmh Jan 11 '21 at 00:51
  • @MichaelMorrow Oh wow. Thank you for this. Makes studying easier. I would still like to know if my answer is a way to go or if it is correct. – Norman Contreras Jan 11 '21 at 00:58
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    @NormanContreras: Something similar to your approach will work, but what you have does not. You’ve shown that if $x\in\operatorname{cl}A$ for some $A\subseteq X$, then $f$ is continuous at $x$. To complete the argument, you must show that each $x\in X$ is in the closure of some subset of $X$. Of course this is trivial, since we can take $A=\operatorname{cl}{x}$, but it needs to be done. I’ll write up something along those lines so that you can see how it can be done. – Brian M. Scott Jan 11 '21 at 01:11
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    @GNUSupporter8964民主女神地下教會 Yes someone already posted that one above. – Norman Contreras Jan 11 '21 at 20:16

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Your idea basically amounts to showing that $f$ is continuous at each point of $X$, but to carry it out properly, you really should start with an arbitrary $x\in X$. Now you need an $A\subseteq X$ such that $x\in\operatorname{cl}A$. That’s easy: let $A=\operatorname{cl}\{x\}$. (Of course if $X$ is $T_1$, $A$ will just be the singleton $\{x\}$, but we don’t really care what it is as long as it contains $x$.)

Now suppose that $f$ is not continuous at $x$. Then there is an open nbhd $U$ of $f(x)$ such that $f[V]\nsubseteq U$ for every open nbhd $V$ of $x$. Thus, for each open nbhd $V$ of $x$ there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $S=f^{-1}[Y\setminus U]$. Then for each open nbhd $V$ of $x$ we have $x_V\in V\cap S$, so $x\in\operatorname{cl}S$, and therefore

$$f(x)\in\operatorname{cl}f[S]\subseteq Y\setminus U\,,$$

which is absurd, since $f(x)\in U$. Thus, $f$ is continuous at $x$ and therefore is continuous.

And when we get to this point, we see that it was really $S$ that we wanted: we never actually used $A$ at all.

Brian M. Scott
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