Vrftjkry

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 $Asubseteq f^{-1}(f(A))subseteq f^{-1}overline{f(A)}$, we conclude that $overline{A}subseteq f^{-1}(overline{f(A)})$, and therefore $fleft(overline{A}right)subseteq fleft(f^{-1}left(overline{f(A)}right)right)subseteq overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).

Conversely, assume that for every $A$, $fleft(overline{A}right)subseteq overline{f(A)}$. 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, $fleft(overline{X-f^{-1}(V)}right)subseteq overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) subseteq Y-V$, which is closed; so
$$fleft(overline{X-f^{-1}(V)}right)=fleft(overline{f^{-1}(Y-V)}right)subseteqoverline{fleft(f^{-1}(Y-V)right)}subseteq Y-V.$$ Therefore,
$$overline{X-f^{-1}(V)} subseteq f^{-1}(Y-V).$$
Is this sufficient?

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 $x_0in X$ 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 $x_Vin V$ such that $f(x_V)notin U$. Let $$A={x_V:Vtext{ is an open nbhd of }x_0};.$$

Clearly $x_0inoverline{A}$, and $f[A]subseteq Ysetminus U$. Moreover, $Ysetminus U$ is closed, so $overline{f[A]}subseteq Ysetminus U$, and hence

$$y_0in f[overline{A}]subseteqoverline{f[A]}subseteq Ysetminus U;,$$

contradicting the choice of $U$.

Those who like nets may notice that $A$ actually is a net, over the directed set $langlemathscr{N}(x_0),supseteqrangle$, 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 $$mathscr{F}={Vcap f^{-1}[Ysetminus U]:Vinmathscr{N}(x_0)}$$ instead.

We show that $f$ is continuous at each $xin X$. Recall that $f$ is continuous at $x$ if for any open nhood $V$ of $f(x)$, there is an open nhood $U$ of $x$ with $f(U)subseteq V$.

So, let $xin X$ and let $V$ be an open nhood of $f(x)$.
Set $E=Xsetminus f^{-1}(V)$. Noting that $f(E)$ is contained in the closed set $V^C$,
we see that
$$f(overline{E})subseteq overline{f(E)}subseteq V^C; $$
whence $xnotin overline E$ (since $f(x)$ is in $V$).

But then $x$ is in the open set $Xsetminus overline{E}$. Moreover, since $Xsetminus overline{E}subseteq Xsetminus E=f^{-1}(V)$, it follows that $f(Xsetminus overline{E})subseteq V$, as desired.

If $X$ were first countable, we could argue using sequences (where a sequence $(x_n)$ converges to $x$ if for any nhood $U$ of $x$ there is an $N$ so that $x_min U$ for all $mge N$); for in such a space a function is continuous at $x$ if and only if $(f (x_n))$ converges to $f(x)$ whenever $x_n$ converges to $x$.

Indeed, it is easily verified that given $x_nrightarrow x$ and any subsequence $(x_{n_k})$ of $(x_n)$, that the image of this subsequence under $f$ when thought of as a sequence has a subsequence that converges to $f(x)$. Thus every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$, which implies that $(f(x_n))$ converges to $x$.

Of course, $X$ need not be first countable...

Though I think the sequential method above is a bit much here, I wonder if the argument can be suitably modified to arbitrary $X$ by using nets?

Here's one proof of the converse provided $X$ and $Y$ are metric spaces:

Take a limit point $x$ of $A$. Then because $f(overline{A}) subseteq overline{f(A)}$, we have that $f(x)$ is a limit point of $f(A)$.

Because $x$ is a limit point of $A$, for every $delta > 0$ there is a point $p in A$ with $|x - p| < delta$. And because $f(x)$ is a limit point of $f(A)$, for every $epsilon > 0$, there is a point $q in f(A)$ with $|f(x) - f(p)| < epsilon$.

We thus have that for every point $x in A$ with $|x - p| < delta$, in fact $|f(x) - f(p)| < epsilon$, so $f$ must be continuous.

A proof using nets:

Suppose $yin f(overline{A})backslash overline{fA}$. Taking $xin overline{A}$ with $y=f(x)$; there is a net $x_iin A$ converging to $x$. Now $f(x_i)$ is a net in $f(A)$ which does not converge to $f(x)$ [for this would imply $f(x)in overline{fA}$]. Thus $f$ is not continuous.

The assertion is equivalent to:

$overline{A}subseteq f^{-1}(overline{f(A)})$

So, the assertion follows from:

$overline{A}subseteqoverline{f^{-1}(f(A))}subseteqoverline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$

1. Inclusion: $Asubseteq f^{-1}(f(A)) Rightarrow overline{A}subseteqoverline{f^{-1}(f(A))}$


2. Inclusion: $f(A)subseteqoverline{f(A)} Rightarrow f^{-1}(f(A))subseteq f^{-1}(overline{f(A)}) Rightarrow overline{f^{-1}(f(A))}subseteq overline{f^{-1}(overline{f(A)})}$


3. Equality: $overline{f(A)} text{ closed} Rightarrow f^{-1}(overline{f(A)}) text{ closed} Rightarrow overline{f^{-1}(overline{f(A)})}=f^{-1}(overline{f(A)})$

The converse assertion is equivalent to:

$overline{B}=B Rightarrow overline{f^{-1}(B)}=f^{-1}(B)$

So, the converse assertion follows from:

$f^{-1}(B)subseteqoverline{f^{-1}(B)}subseteq f^{-1}(f(overline{f^{-1}(B)}))subseteq f^{-1}(overline{f(f^{-1}(B))}) subseteq f^{-1}(overline{B}) =f^{-1}(B)$

That gives:

$f^{-1}(B)=overline{f^{-1}(B)}$

1. Inclusion: $Asubseteq overline{A} text{ in general}$


2. Inclusion: $Asubseteq f^{-1}(f(A)) text{ in general}$


3. Inclusion: $f(overline{A})subseteq overline{f(A)} text{ by assumption}$


4. Inclusion: $f(f^{-1}(B))subseteq B text{ in general} Rightarrow overline{f(f^{-1}(B))}subseteq overline{B} Rightarrow f^{-1}(overline{f(f^{-1}(B))})subseteq f^{-1}(overline{B})$


5. Equality: $overline{B}=B Rightarrow f^{-1}(overline{B})=f^{-1}(B)$