0

I have the following exercise and I don't completely understand some things in the answer given by the textbook:

Exercise: Show that $f:(X,\mathcal{T}_X)\to (Y,\mathcal{T}_Y)$ is continuous if and only if $f(\overline{A})\subset \overline{f(A)}$ for every $A\subset X$. (Hint: for 'if', show that whenever $F$ closed in $Y$ and $A:=f^{-1}(F)$, then $\overline{A}\subset A$. Justify why this is sufficient.).

The answer is the following:

If f is continuous then $f^{-1}\overline{(f(A))}$ is a closed set in $X$ that contains $A$, so contains $A$:

$\overline{A}\subset f^{-1}\overline{(f(A))}.$

Applying $f$ to both sides shows that $f(\overline{A})\subset \overline{f(A)}$.

For the "if" proof i'm struggling to understand why taking a closed $F\in Y$ and then proving that $A=f^{-1}(F)$ is closed is sufficient to show that for every $A$ the function is continuous. What if $A$ is neither closed or open i.e it is just a set $A\subset X$? How do we prove it for that case?

The "if" proof:

Now suppose that $f(\overline{A})\subset \overline{f(A)}$. Take $F$ closed in $Y$ and set $A=f^{-1}(F)$; we need to show that $A$ is closed.

To do this we show that any $x\in\overline{A}$ is in $A$: this is sufficient, since we always have $A\subset \overline{A}$, so showing that $\overline{A}\subset A$ implies equality, and $A$ is closed if and only if $A=\overline{A}$.

Noting that $f(A)=f(f^{-1}(F))\subset F$ and using our assumption we have $$f(x)\in f(\overline{A})\in\overline{f(A)}\subset \overline{F}=F,$$

and so $x\in f^{-1}(F)=A$, i.e. $\overline{A}\subset A$.

I'm not quite sure why $x\in f^{-1}(F)=A$ implies $\overline{A}\subset A$. Is it because $f(x)\in f(\overline{A})$ and hence $x\in \overline{A}$. And since $f(\overline{A})\subset F$ then this means that $f(x)$ belonging to $f(\overline{A})$ means it belongs to $F$ and hence every $x\in \overline{A}$ also belongs to $A$?

  • 1
    First question: the definition of continuity is that the preimage of any open set is open; this is equivalent to insisting the preimage of any closed set is closed (try using $F=Y\setminus U$ for an open $U$). Second question: you've taken $x\in \overline{A}$ and then you show that $x\in A$, so $\overline{A}\subseteq A$. The assumption of $x\in\overline{A}$ is what's used to say $f(x)\in f(\overline{A})$. – Dave Jul 20 '21 at 14:22
  • I spell out the same argument here. – Henno Brandsma Jul 20 '21 at 21:15

1 Answers1

0

If $f$ is continuous then by definition we know that for every open subset $O$ of $Y$, $f^{-1}[O]$ is open in $X$.

But this also implies that if $F$ is closed in $Y$, then $f^{-1}[F]$ is closed in $X$ as well: $F$ closed implies $Y\setminus F$ open and so by continuity $f^{-1}[Y \setminus F] = X\setminus f^{-1}[F]$ is open which implies (by the last identity) that $f^{-1}[F]$ is closed in $X$.

And by doing the same proof, we see that we can define $f$ continuous as this inverse preservation of closed sets as well: knowing that we get the open fact as well, doing complements and using the same basic set theory facts.

By that last observation we see it's enough to show $A:= f^{-1}[F]$ to be closed just assuming $F$ is closed. And a set $A$ is closed iff $\overline{A} \subseteq A$ because the other inclusion is always true and a set is closed iff it equals its closure.

Henno Brandsma
  • 242,131