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I have found plenty of examples for $f[\overline{A}] \subsetneq \overline{f[A]}$ where $f: X \to Y$ is continuous: Examples for $f(\bar A)\subsetneq\overline{f(A)}$ with $f$ being continuous, Example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$. But every single one of them is such that $A$ is already closed, so $A = \overline{A}$, and so these are just examples of $f[A] \subsetneq \overline{f[A]}$, i.e. of continuous maps that are not closed.

What would be an example where $A$ is not closed?

Anakhand
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    Take any example where $A$ is closed, and remove a point from it, i.e. consider $A\setminus{a}$ for some $a\in A$. I think you will very quickly find what you want. – imtrying46 Sep 25 '24 at 12:20
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    How about just using the first example in this answer but switching $A$ from $\mathbb{R}$ to $\mathbb{Q}$? https://math.stackexchange.com/a/2686321/37122 – Benjamin Dickman Sep 25 '24 at 12:22
  • In the second link you gave, the first answer (with the indiscrete toology on $Y$) gives examples where this is true of almost any $A$, including non-closed $A$. – Simon Pitte Sep 25 '24 at 12:26
  • @SimonPitte I'm not sure I'm seeing the same first answer as you are, could you add the direct link? – Anakhand Sep 25 '24 at 12:28
  • @Anakhand See the answer I just posted detailing what I meant. – Simon Pitte Sep 26 '24 at 13:55

2 Answers2

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Take $f\colon\Bbb R\longrightarrow\Bbb R$ defined by $f(x)=\frac1{x^2+1}$ and $A=\Bbb Q$. Then $A$ is not closed,$$f\left(\overline A\right)=f\bigl(\Bbb R\bigr)=(0,1],$$and $\overline{f(A)}=[0,1]$.

José Carlos Santos
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Take $f:X\rightarrow Y$, and equip $Y$ with the indiscrete, or trivial, topology, that is the only open sets are $\varnothing$ and $Y$. Then $f$ is automatically continuous, and for any $A\subseteq X$ not empty, we have $\overline {f(A)}=Y$. However, so long as $f$ is not surjective, any $A$ will satisfy $f(\overline{A})\neq Y$.

Simon Pitte
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