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Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces and let $f: X \to Y $ be a continuous function. For $A \subset X$ prove that $$ f(\overline{A}) \subset \overline{f(A)}$$

Definition: A mapping $f:X \mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $\epsilon > 0$ there is a $\delta >0$ such that $$y \in X, d_X(x,y) < \delta \implies d_Y(f(x),f(y)) < \epsilon$$

The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .

$\overline{A}$ := closure of $A$

I think I need to show that for every $x \in f(\overline{A})$, $x$ is also element of $\overline{f(A)}$

But I couldn't figure out how to derive it from the definition and connect it with closure

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  • $\begingroup$ Let $y \in f[\overline{A}]$ so $y=f(x)$ with $x \in \overline{A}$. Let $r>0$ be given and we want to show that $B(y,r)$ intersects $f[A]$, so that $y$ is in the closure of that set. Apply continuity to $x,y,\varepsilon=r$ and get a $\delta>0$ as in the definition. Now use that $x \in \overline{A}$ so there is some $x'\in B(x,\delta) \cap A$, and continuity now tells us that $f(x') \in B(y,r) \cap f[A]$. Done. $\endgroup$
    – Henno Brandsma
    Dec 27, 2018 at 22:51
  • $\begingroup$ Note that the duplicate question answers the reverse implication, and in general spaces. Not this forward one; so it's not really a duplicate. $\endgroup$
    – Henno Brandsma
    Dec 27, 2018 at 22:53

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