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For a continuous function $f:X\longrightarrow Y$ of topological spaces with subsets $A$ and $B$ of $X$ satisfying $\overline{A}=\overline{B}$, is it true that $\overline{f(A)}=\overline{f(B)}$? Intuitively it seems so, but if someone could walk me through how this could be proved, I would be very grateful, regards.

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up vote 5 down vote accepted

Pick $y\in \overline {f(A)}$ and a let $U$ be an open neighbourhood of $y$. We must show that $U\cap f(B)\ne \emptyset$. Since $f(A)\cap U\ne\emptyset$ there is a $x_0\in A$ so that $f(x_0)\in U$. Now, $x_0\in \overline B$ and $f^{-1}(U)$ is an open neighbourhood of $x_0$. Hence there exists $x_1\in B\cap f^{-1}(U)$. Therefore $f(x_1)\in U\cap f(b)$ which implies that $y\in \overline {f(B)}$.

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and the reverse inclusion holds by symmetry. – Henno Brandsma Feb 13 '12 at 5:18

You might also observe that

PROPOSITION $f:(X,\mathcal I)\to (Y,\mathcal L)$ is continuous if and only if for each $A\subseteq X$, $f(\bar A)\subseteq \overline{f(A)}$

PROOF Suppose that $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.

Conversely, suppose that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$. Let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.

NOTE From the above proposition, we can see the condition $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ can be taken as the definition of continuity in closure spaces.

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This certainly deserves a +1. – 1015 Feb 28 '13 at 17:52
@julien Thank you. I think it was worth adding. There are around 5 equivalent characterizations of continuity, and this one is little known. – Pedro Tamaroff Feb 28 '13 at 18:09

Here is a hint/partial answer. Because $\bar{A}=\bar{B}$, we have $A \subseteq \bar{B}$ and $B \subseteq \bar{A}$. Hence $\overline{f(A)} \subseteq \overline{f(\bar{B})}$ and $\overline{f(B)} \subseteq \overline{f(\bar{A})}$. If $\overline{f(\bar{B})} = \overline{f(B)}$ and $\overline{f(\bar{A})} = \overline{f(A)}$ then we are done.

Therefore, it is sufficient to prove that, for every set $C \subseteq X$, $\overline{f(C)} = \overline{f(\bar{C})}$. The left to right inclusion for these two sets is immediate because $C \subseteq \bar{C}$. The right to left inclusion $\overline{f(\bar{C})} \subseteq \overline{f(C)}$ is all that's left.

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