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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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).

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

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