Let $f : (X, d_X) \to (Y, d_Y)$ be continuous. Assume $A, B \subseteq X$ satisfy $\bar{A} = \bar{B}$. Show that $\overline{f[A]} = \overline{f[B]} $

Question

Let $f : (X, d_X) \to (Y, d_Y)$ be a continuou map between metric spaces. Assume $A, B \subseteq X$ satisfy $\bar{A} = \bar{B}$. Show that $\overline{f[A]} = \overline{f[B]} $

This is what I did so far

Attempt at proof: Let $y \in \overline{f[A]}$, then either $y = f(a)$ for some $a \in A$ or $y \in \overline{f[A]}\setminus f[A]$.

In the first case since $\overline{A} =\overline{B}$ we have $A \subseteq \bar{B}$ hence $f(a) \in f[\bar{B}] \subseteq \overline{f[B]}$ by continuity. So $y \in \overline{f[B]}$ in the first case.

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In the second case I tried to show that if $y$ is a limit point of $f[A]$, then $y$ is a limit point of $f[B]$ but I got stuck.

Is there an easy (sequence free) way to do this without having to view limit points as the limit of sequences?

If so then this must generalize to arbitrary topological spaces correct?

Answer 1

Since $f$ is continuous map then $f^{-1}(\overline{f(A)})$ must be closed in $X$. But $f(A)\subseteq \overline{f(A)}$ then $A\subseteq f^{-1}(\overline{f(A)})$. However $\overline{A}$ is the smallest closed set containing $A$ hence $\overline{A}\subseteq f^{-1}(\overline{f(A)})$ which using the assumption $\overline{A}=\overline{B}$ yields $B\subseteq\overline{B}\subseteq f^{-1}(\overline{f(A)})$. This implies $f(B)\subseteq \overline{f(A)}$. But $\overline{f(B)}$ is the smallest closed set containing $f(B)$ hence $\overline{f(B)}\subseteq \overline{f(A)}$. Analogue argument for the other inclusion $\overline{f(A)}\subseteq \overline{f(B)}$.

Answer 2

In general, we have the following. If $f: X \to Y$ is a continuous map of topological spaces, and $A\subset X$ then $\overline{f(\overline{A})}= \overline{f(A)}$. Your claim follows immediately from this. You can check out this question:

Closure of continuous image of closure