2
$\begingroup$

If $f(\overline A)\subseteq\overline{f(A)}$ for all $A\subset{\rm dom}(f)$ then $f$ is continuous.

I know there are others questions about this topic but I want to know what more I need to conclude my proof, if possible. The way Im trying to prove this is different to what I see in other answers about this topic. The context of this proof is

$$f:X\to Y$$

where $X$ and $Y$ are metric spaces.

Then to proof that if $f(\overline A)\subseteq\overline{f(A)}$ then $f$ is continuous I will reason with convergent sequences.

We have that for any sequence $(x_n)\to x$ with $x_n\in A$ for all $n\in\Bbb N$ then $x\in\overline A$.

Then observe that the statement $f(\overline A)\subseteq\overline{f(A)}$ implies that if $f(x_n)\in f(A)$ then $f(x)\in\overline{f(A)}$, i.e. the image of the limit of any convergent sequence in $A$ is defined in the closure of the image of $A$, in other words

$$f(\lim x_n)=f(x)\in\overline{f(A)}$$

But I dont know if I can conclude that

$$\lim f(x_n)=f(x)$$

what is the definition of continuity on metric spaces. In other words: what I need to conclude

$$\lim f(x_n)=f(\lim x_n)$$

in this context?

$\endgroup$
  • $\begingroup$ You haven't made use of the fact that the subset condition holds for all $A \subset dom(f)$. There may be some mileage in considering a subset consisting just of the points$(x)_n$. $\endgroup$ – Tom Collinge Nov 21 '16 at 13:58
4
$\begingroup$

The proof in topological spaces:

Assume that $f$ is not continuous, that is there exists some open set $V\subset Y$ with $U:=f^{-1}(V)$ is not open. Let $A = X\setminus U$. Notice that for every $x\in X$ we have $$ x \in A = X\setminus f^{-1}(V) \iff x\notin f^{-1}(V) \iff f(x) \notin V,$$ that is $f(A) \subseteq Y\setminus V$.

Now, as $U$ is not open, there exists a $x\in U \cap \bar A$ and we have

  1. $x\in U = f^{-1}(V) \iff f(x) \in V$ and
  2. $x\in \bar A \implies f(x) \in f(\bar A) \subseteq \overline{f(A)} \subseteq Y \setminus V$,

which is a contradiction.

Old answer:

Proof only for metric spaces:

Assume the converse: Let $x_n\to x$ and $f(x_n)\not\to f(x)$. By going to a subsequence we may assume that there exists a $\epsilon > 0$ such that $|f(x_n) - f(x)| \ge \epsilon$ holds for all $n$. On other hand we have $$ f(x) \in \overline{\{ f(x_n) \mid n\in \mathbb N \}}. $$ That is, there exists a sequence $f(x_{n_k}) \to f(x)$, which is a contradiction.

$\endgroup$
  • $\begingroup$ Good answer: I added some notes. $\endgroup$ – Tom Collinge Nov 22 '16 at 14:20
  • $\begingroup$ To me, there is a gap in your Old answer. I cannot see why you can assume that one can always find a subsequence such that for all $n$, $|f(x)-f(a)|>\epsilon>0$. Can you clarify that please? It is easy to imagine that for some $n$, $|f(x)-f(a)|>\epsilon$, but not for all $n$. $\endgroup$ – Drake Marquis Feb 6 '18 at 7:01
  • $\begingroup$ @DrakeMarquis "by going to a subsequence" ... that subsequence must be infinite, otherwise, $f(x_n)$ converges. $\endgroup$ – user251257 Feb 6 '18 at 12:59
  • $\begingroup$ Would you please explicitly construct such a subsequence? $\endgroup$ – Drake Marquis Feb 7 '18 at 5:37
  • $\begingroup$ @DrakeMarquis you should ask a new question. Similar questions are probably answered before. $\endgroup$ – user251257 Feb 7 '18 at 9:07
3
$\begingroup$

This is too long for a comment, but hopefully a useful addition.

The topological proof given by user251257 is neat, but perhaps worth some expansion.

If $U$ is not open then $X \setminus U$ is not closed and therefore doesn't contain all its limit points. So there is a limit point $x$ of $X \setminus U$ in $X \setminus (X \setminus U) = U$.

$x $ and other limit points of $X \setminus U$ are in the closure of $X\setminus U $, so as stated there is $x \in U \cap \bar A$ and it follows (1) $f(x) \in V$.

For (2), $ \overline{f(A)} $, the closure of $f(A)$ is a subset of every closed set that contains $f(A)$; since $V$ is open then $Y \setminus V$ is closed, and as shown initially $f(A) \subset Y \setminus V$: therefore $Y \setminus V $ is a closed set containing $f(A)$ and therefore contains $ \overline{f(A)} $.

(Since it took me a while to work this out I thought I'd share it).

$\endgroup$
  • $\begingroup$ Thx. It was late and I was lazy :D $\endgroup$ – user251257 Nov 22 '16 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.