Continuity via Closure Operator

Question

I'm having a rather simple question:

Lets say a function preserves neighborhoods iff: $N\in\mathcal{N}_x \Rightarrow f^{-1}(N)\in\mathcal{M}_x$
and a function preserves closeness iff: $x\parallel A \Rightarrow f(x)\parallel f(A)$

I want to show that, in fact, these are equivalent.
So far the second property is the same as: $f(\overline{A})\subseteq \overline{f(A)}$

I already passed a proof that a function preserves neighborhoods iff it is continuous in the usual sense: $V\in\mathcal{T} \Rightarrow f^{-1}(V)\in\mathcal{S}$

...while $x\parallel A$ is meant to mean $x\in\overline{A}$.

what does "$x\parallel A$" mean? – Stefan Hamcke Jan 22 '14 at 17:56 — Stefan Hamcke, Jan 22 '14 at 17:56
Oh, I meant that it is in the closure: $x\in\overline{A}$ – C-star-W-star Jan 22 '14 at 18:07 — C-star-W-star, Jan 22 '14 at 18:07
I have never encountered that notation. – Stefan Hamcke Jan 22 '14 at 18:08 — Stefan Hamcke, Jan 22 '14 at 18:08

Stefan Hamcke · Accepted Answer · 2014-01-22T21:44:53.350

2

For the $\implies$ direction, I suggest a proof by contradiction. Assume that $x\in\bar A$ but $f(x)\notin\overline{f(A)}$. Then there is a neighborhood $V$ of $f(x)$ disjoint from $f(A)$. This is equivalent to $f^{-1}(V)\cap A=\emptyset$, contradicting $x\in\overline A$, as $f^{-1}(V)$ is a neighborhood by hypothesis.
For the other direction, assume that $V$ is a neighborhood of $f(x)$, but no neighborhood of $x$ has an image in $V$. Then $x$ is in the closure of $f^{-1}(Y\setminus V)$, thus $f(x)$ is in the closure of $f(f^{-1}(Y\setminus V))$ by hypothesis, hence it is also in the closure of the larger set __ ... (I'll leave it to you to finish the argument.)

This actually proves a stronger statement, as it works for a fixed point $x$ while $A$ ranges over the subsets of $X$. So it shows that continuity at a point $x$ is equivalent to the implication
"$x\in\overline A\implies f(x)\in\overline{f(A)}$ for all $A\subset X$"

edited Jan 22 '14 at 21:44

answered Jan 22 '14 at 18:06

Stefan Hamcke

27,733

1

I haven't dived into this, but the implication that you mention in the last line is trivially satisfied. You probably forgot some bars. – drhab Jan 22 '14 at 19:07
^^ ...ok let me think... – C-star-W-star Jan 22 '14 at 19:35
@drhab: Thanks :) – Stefan Hamcke Jan 22 '14 at 21:31
@Freeze_S: I didn't mean the "fill in the details" literally :-) Sorry, if it appeared like inviting you to edit the answer into my post. If you'd like to get feedback, it's generally better to leave a comment instead of editing the post. But you were right about the "$x\in\overline{f^{-1}(Y\setminus V)}$. – Stefan Hamcke Jan 22 '14 at 21:35
Haha ^^ I thought it's a riddle – C-star-W-star Jan 23 '14 at 00:21
I just rejected the "unfilling" edit (prima facie judgement). Better let Stefan do any editing he deems appropriate. – ccorn Jan 23 '14 at 00:27
I don't mind ;) ...ok got it - I was thinking wrong. – C-star-W-star Jan 23 '14 at 00:31

score 1 · Answer 2 · answered Jan 22 '14 at 18:42

...ok, sry, I got it by myself ...thanks, Stefan, anyway! #thumbs-up# =)

The assertion is equivalent to:
$\overline{A}\subseteq f^{-1}(\overline{f(A)})$
So, the assertion follows from:
$\overline{A}\subseteq\overline{f^{-1}(f(A))}\subseteq\overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$

Inclusion: $A\subseteq f^{-1}(f(A)) \Rightarrow \overline{A}\subseteq\overline{f^{-1}(f(A))}$
Inclusion: $f(A)\subseteq\overline{f(A)} \Rightarrow f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)}) \Rightarrow \overline{f^{-1}(f(A))}\subseteq \overline{f^{-1}(\overline{f(A)})}$
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:
$f^{-1}(B)\subseteq\overline{f^{-1}(B)}\subseteq f^{-1}(f(\overline{f^{-1}(B)}))\subseteq f^{-1}(\overline{f(f^{-1}(B))}) \subseteq f^{-1}(\overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=\overline{f^{-1}(B)}$

Inclusion: $A\subseteq \overline{A} \text{ in general}$
Inclusion: $A\subseteq f^{-1}(f(A)) \text{ in general}$
Inclusion: $f(\overline{A})\subseteq \overline{f(A)} \text{ by assumption}$
Inclusion: $f(f^{-1}(B))\subseteq B \text{ in general} \Rightarrow \overline{f(f^{-1}(B))}\subseteq \overline{B} \Rightarrow f^{-1}(\overline{f(f^{-1}(B))})\subseteq f^{-1}(\overline{B})$
Equality: $\overline{B}=B \Rightarrow f^{-1}(\overline{B})=f^{-1}(B)$

I love this proof. It's only downside is that it does not talk about continuity at a point. — Carsten Führmann, Jun 13 '19 at 15:43
That's true. I guess one may (should) formulate this rather pointwise in the sense $x\parallel A\implies f(x)\parallel f(A)$ for every $A\subset X$ if and only if $N\in\mathcal{N}_{f(x)}\implies f^{-1}N\in\mathcal{M}_x$. — C-star-W-star, Jun 15 '19 at 08:26

user68061 · Answer 3 · 2014-01-22T18:25:27.683

So I misunderstood the question, here is an update: if $f$ is continuous then $f^{-1}(X\setminus \overline{f(A)})$ is open. So its complement is closed and contains $A$. Thus it contains $\overline{A}$. So $f(\overline{A}) \subset \overline{f(A)}$.

Other direction: Let $U$ be an open set and $V=f^{-1}(U)$. We want to show that $V$ is open. Take $B=X \setminus U, A=f^{-1}(B)$. Then $\overline{f(A)} \subset B$ so $f(\overline{A}) \cap U= \emptyset$ . Thus $\overline{A}=A$ and $V=X \setminus A$ is open.

Previous version:

The first property is, indeed, continuity, but the second is not. Functions, that satisfy this property are called closed. For example, map the open unit interval to the circle by ``folding'' the ends together. Then the image of the interval itself (it is closed in induced topology) is not closed in the topology of circle, however the map is continuous.

See: http://en.wikipedia.org/wiki/Open_and_closed_maps

Yeah, I know what you mean, but I'm asking wether the closure is preserved - that is slightly different... — C-star-W-star, Jan 22 '14 at 18:09

Continuity via Closure Operator

3 Answers3