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I already know of this question ($f$ is continuous and closed $\Longleftrightarrow \overline{f(E)} = f(\overline{E})$ for all $E \subseteq M$) which is quite the same as mine.

But I don't understand some chunks of the proof.

First, let $x \in \overline{A}$, we can build a sequence $u_n \in A^{\mathbb{N}}$ which converges to $x \in \overline{A}$.

Also, we already know by definition that $f(x) \in f(\overline{A})$.

Then, as $f$ is continuous, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(x)$, so we deduce $f(x) \in \overline{f(A)}$ (I don't fully understand why we can deduce this.)

Now, we have $f(\overline{A}) \subset \overline{f(A)}$.

Finally, as $f(\overline{A})$ is a closed set because $f$ is a closed map and $\overline{A}$ is a closed set by definition.

We can deduce that $f(A) \subset f(\overline{A})$ and as $f(\overline{A})$ is closed, $f(A)$ being included in it indicate that we also have smallest closed set containing $f(A)$ in $f(\overline{A})$, at least, I understand this. As a result, we can deduce that $\overline{f(A)} \subset f(\overline{A})$, but why?

I am a bit lost and new to closed / open sets, sorry for the formatting.

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  • $\begingroup$ Okay, the fact that $\overline{A}$ is a closed set is by definition, sorry for this stupid question. $\endgroup$ – Raito Aug 10 '17 at 23:19
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    $\begingroup$ Also if $A \subseteq B$ and $B$ is closed, then $\overline{A} \subseteq \overline{B} = B$, or use minimality directly. We use properties like $A \subseteq \overline{A}$, $A \subset B$ implies $\overline{A} \subseteq \overline{B}$ and $\overline{\overline{A}} =\overline{A}$ etc. $\endgroup$ – Henno Brandsma Aug 11 '17 at 5:19
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The topological def'n is that $f:X\to Y$ is continuous iff $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y.$ Regardless of whether the topologies can be defined in terms of sequences. There are many consequences of the continuity of $f$ that also imply the continuity of $f,$ and could be taken as alternate, equivalent def'ns. One of these is : $$f(Cl_X(A))\subset Cl_Y(f(A)) \;\text {for all } A\subset X.$$

(i). Suppose $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y$. Then for $A\subset X,$ the set $C=Y \backslash Cl_Y(f(A))$ is open in $Y,$ so $f^{-1}C$ is open in $X,$ and $(f^{-1}C)\cap A=\phi.$

Now for any open set $P,$ if $P\cap Q=\phi$ then $P\cap \bar Q=\phi.$ So $(f^{-1}C)\cap \bar A=\phi.$ Therefore $f(\bar A)\subset Y \backslash C=Cl_Y(f(A)).$

(ii). I will leave the converse of (i) to you.

By (i), if $f:E\to F$ is a continuous closed map and $A\subset E$ then $$f(\bar A)\subset \overline {f(A)}\subset \overline {f(\bar A)} \;\text { by continuity of } f$$ $$\text { and }\quad f(\bar A)= \overline {f(\bar A)}\; \text { by closedness of } f$$ $$\text {so }\quad \overline {f(A)}=f(\bar A).$$

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  • $\begingroup$ Interesting, but I don't fully understand why $(f^{-1}C \cap A = \emptyset)$, and I am unable to prove exactly why $\forall \text{ open set in X } P, \forall Q \in P(X), P \cap Q = \emptyset \implies P \cap \overline{Q} = \emptyset$, neither could I found a counter-example… I tried to use balls to understand intuitively the result, but failed to do so. $\endgroup$ – Raito Aug 11 '17 at 13:53
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    $\begingroup$ (1). $ x\in A\implies$ $ f(x)\in f(A)\subset Cl_Y(f(A)$ $)\implies f(x)\not \in C$.... (by def;n of $C$)...$\implies x\not \in f^{-1}C$........This is easier to see pictorially .....(2). If $x\in \overline Q$ then every nbhd of $x$ has non-empty intersection with $Q.$ If $x\in P$ where $P$ is open and $P\cap Q=\phi$ then $ P$ is a nbhd of $x$ that has empty intersection with $Q$ so $x\not \in \overline Q.$.... You can also say that $Q \subset X$ \ $P,$ and $X$ \ $P$ is closed, so $Cl(Q) \subset Cl( X$ \ $P)=X$ \ $P. $ $\endgroup$ – DanielWainfleet Aug 11 '17 at 16:01
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Then, as $f$ is continuous, the sequence $(f(u_n))_{n \in \mathbb{N}}$ converges to $f(x)$, so we deduce $f(x) \in \overline{f(A)}$ (I don't fully understand why we can deduce this.)

Let $y = f(x)$ and $v_n = f(u_n)$ for each $n$, then what is stated there is that the sequence $v_n$ converges to $y$ (by definition of continuity). On the other hand, each $v_n$ is in $f(A)$ because each $u_n$ is in $A$, and the limit of any sequence in $f(A)$ lies in $\overline{f(A)}$ (can you see why is this the case?), therefore $y$ itself, being the limit of elements in $f(A)$, belongs to $\overline{f(A)}$. In other words, $f(x)\in \overline{f(A)}$.

We can deduce that $f(A) \subset f(\overline{A})$ and as $f(\overline{A})$ is closed, $f(A)$ being included in it indicate that we also have smallest closed set containing $f(A)$ in $f(\overline{A})$, at least, I understand this. As a result, we can deduce that $\overline{f(A)} \subset f(\overline{A})$, but why?

Did you understand this claim: we also have smallest closed set containing $f(A)$ in $f(\overline{A})$? If you do, then you're done because by definition (or, as a property depending on your definition of closure) the set $\overline{f(A)}$ is precisely "smallest closed set containing $f(A)$".

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$f: X \to Y$ is continuous iff $$\forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$$

See this question and its answers.

$f: X \to Y$ is closed iff $$\forall A \subseteq X: \overline{f[A]} \subseteq f[\overline{A}]$$

If $f$ is closed then pick any $A \subseteq X$ then $f[A] \subseteq f[\overline{A}]$ as $A \subseteq \overline{A}$. But $f[\overline{A}]$ is closed as the image of a closed set under $f$ so $\overline{f[A]} \subseteq \overline{f[\overline{A}]} = f[\overline{A}]$.

Now suppose the inclusion property holds and let $A$ be closed in $X$. Then $A = \overline{A}$ and we get $(f[A] \subseteq) \overline{f[A]} \subseteq f[A]$, so $f[A]$ equals its closure and so $f[A]$ is closed.

From these two equivalences we get the characterisation of closed and continuous functions.

Nothing we do here uses metrics, just general properties of closure and continuity.

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$\overline {A} $ is closed, not because $A $ is open but if $(a_n) $ is a sequence of elements from $A $, which converges to $a $, then

$$\forall \epsilon>0, \exists N\in \Bbb N \;\;:\;\; d (a,a_N)<\epsilon $$ and this means that $a\in \overline {A} $.

For example $A=(0,1] $ is not open but $\overline {A}=[0,1] $ is closed.

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  • $\begingroup$ This does not explain why $f(x) \in \overline{f(A)}$, nor $\overline{f(A)} \subset f(\overline{A})$. $\endgroup$ – Raito Aug 10 '17 at 23:18
  • $\begingroup$ @LionelRicci Correct me if I'm wrong, let be $S$ a closed set, let be $S'$ the set of its limit points, then $\overline{S} = S \cup S'$, but as $\overline{S} = S$, we conclude that $S' \subset S$ ? $\endgroup$ – Raito Aug 10 '17 at 23:37

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