# If the closures of two subsets of a topological space are equal, are the closures of the images of the two subsets under a continuous map also equal?

Suppose $A$ and $B$ are subsets of a topological space $X$ such that $\newcommand{cl}{\operatorname{cl}}\cl(A) = \cl(B)$.

Let $f\colon X\to Y$ be a continuous map of topological spaces.

Does that mean that $\cl(f(A)) = \cl(f(B))$?

-

## 3 Answers

Let $S\subset Y$ be a closed set containing $f(A)$. Since $f$ is continuous, $f^{-1}(S)$ is also closed. Moreover, we have $$f^{-1}(S)\supset f^{-1}(f(A))\supset A.$$ Therefore, we see that $\cl(B)=\cl(A)\subset f^{-1}(S)$. In particular, we have $B\subset f^{-1}(S)$. Applying $f$, we get $$f(B)\subset f(f^{-1}(S))\subset S.$$

Hence, any closed set containing $f(A)$ also contains $f(B)$. By symmetry, any closed set containing $f(B)$ also contains $f(A)$. Therefore, we conclude that $\cl(f(A))=\cl(f(B))$.

-

$\newcommand{\ol}[1]{\overline{#1}}\newcommand{\Obr}[2]{{#1}[{#2}]}$A function between two topological spaces $X$ and $Y$ is continuous if and only if $\Obr f{\ol A}\subseteq\ol{\Obr fA}$ holds for every $A\subseteq X$, see e.g. Wikipedia.

Now if $f$ is continuous and $M\subseteq N\subseteq \ol M$, then $\ol{\Obr fM}=\ol{\Obr fN}$. To see that this holds, just notice that $\Obr fM \subseteq \Obr fN \subseteq \Obr f{\ol M} \subseteq \ol{\Obr fM}$. If we apply the closure to $\Obr fM \subseteq \Obr fN \subseteq \ol{\Obr fM}$, we get $\ol{\Obr fM} \subseteq \ol{\Obr fN} \subseteq \ol{\Obr fM}$.

Now if we use the above observation for $M=A$ and $N=\ol B$, we get $\ol{\Obr fA}=\ol{\Obr f{\ol B}}\supseteq \ol{\Obr fB}$. By symmetry, the opposite inclusion must be true, too. Hence $\ol{\Obr fA}=\ol{\Obr fB}$.

-

The answer is yes. (Thanks Dejan Gove for pointing out that my first attempt, which tried to prove the answer is no, was wrong!)

By symmetry it suffices to prove $\overline{fA}\subset \overline{fB}$.

Let $y\in \overline{fA}$; take a net $a_i\in A$ with $y=\lim_i f(a_i)$.

Then $a_i\in A\subset\overline{A}=\overline{B}$, hence $f(a_i)\in f(\overline{B})\subset \overline{fB}$ (the last inclusion holds by continuity).

Since the closure is closed, $y=\lim_i f(a_i)\in \overline{fB}$.

-