A map is continuous if and only if for every set, the image of closure is contained in the closure of image

As a part of self study, I am trying to prove the following statement:

Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$.

Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me?

Solution 1:

Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:

If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.

Now using our closure property for $D$: $$f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C} = C,$$ as $C$ is closed.

This means that $\overline{D} \subseteq f^{-1}[C] = D$, making $D$ closed, as required.

Solution 2:

If $f$ is continuous, then $f^{-1}(Y-\overline{f(A)})$ is open; since $f^{-1}(Y-\overline{f(A)}) = X -f^{-1}(\overline{f(A)})$, then $f^{-1}(\overline{f(A)})$ is closed; since $A\subseteq f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)})$, we conclude that $\overline{A}\subseteq f^{-1}(\overline{f(A)})$, and therefore $f\left(\overline{A}\right)\subseteq f\left(f^{-1}\left(\overline{f(A)}\right)\right)\subseteq \overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).

Conversely, assume that for every $A$, $f\left(\overline{A}\right)\subseteq \overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=\overline{X-f^{-1}(V)}$.

By assumption, $f\left(\overline{X-f^{-1}(V)}\right)\subseteq \overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) \subseteq Y-V$, which is closed; so $$f\left(\overline{X-f^{-1}(V)}\right)=f\left(\overline{f^{-1}(Y-V)}\right)\subseteq\overline{f\left(f^{-1}(Y-V)\right)}\subseteq Y-V.$$ Therefore, $$\overline{X-f^{-1}(V)} \subseteq f^{-1}(Y-V).$$ Is this sufficient?

Solution 3:

Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.

Suppose that $f$ is not continuous. Then there is some point $x_0\in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $$A=\{x_V:V\text{ is an open nbhd of }x_0\}\;.$$

Clearly $x_0\in\overline{A}$, and $f[A]\subseteq Y\setminus U$. Moreover, $Y\setminus U$ is closed, so $\overline{f[A]}\subseteq Y\setminus U$, and hence

$$y_0\in f[\overline{A}]\subseteq\overline{f[A]}\subseteq Y\setminus U\;,$$

contradicting the choice of $U$.

Those who like nets may notice that $A$ actually is a net, over the directed set $\langle\mathscr{N}(x_0),\supseteq\rangle$, where $\mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)

Those who prefer filters may modify this to consider the filter $$\mathscr{F}=\{V\cap f^{-1}[Y\setminus U]:V\in\mathscr{N}(x_0)\}$$ instead.