$\vert\{\alpha\in I:F_\alpha\cap V\neq\emptyset\}\vert<\aleph_0\implies \bigcup\limits_{\alpha\in I} F_\alpha$ is closed.

Question

Let $X$ be a topological space and $\{F_\alpha:\alpha\in I\}$ a family of closed subsets of $X.$ If for all $x\in X,$ there exists $V\in\mathcal V(X)$ such that $\vert\{\alpha\in I:F_\alpha\cap V\neq\emptyset\}\vert<\aleph_0$, then $\displaystyle\bigcup_{\alpha\in I} F_\alpha$ is closed.

I have no much idea on how to prove it.

I've checked others questions and found only one similar Natural conditions implying the union of closed sets is closed and I think the conversely of Alex Ravinsky's answer has something to do with my statement:

Conversely, if a point $x$ has a neighborhood $O_x$ intersecting only finitely many members of the family $\mathcal F$ then $O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$ is an open neighborhood of the point $x$ disjoint from $F$.

However I could not see how the arguments induce to the desired result $\displaystyle\bigcup_{\alpha\in I} F_\alpha$ is closed.

Can someone please help me with this difficult exercise?

Thanks in advance.

Might want to take a look at this https://math.stackexchange.com/questions/2598645/showing-that-a-locally-finite-collection-preserves-the-closure — R Los, May 29 '19 at 03:33
A locally finite collection is what you have. By definition, a collection of sets of a topological space is locally finite if every point in the space has a neighbourhood which intersects only finitely many elements of such a collection. That's what you wrote in symbols at the beginning of your text. — R Los, May 29 '19 at 03:59
@WilliamElliot borel $\sigma-algebra$ that contains the topology $\tau$ — , May 29 '19 at 04:45

score 2 · Answer 1 · answered May 29 '19 at 07:33

Conversely, if a point $x$ has a neighborhood $O_x$ intersecting only finitely many members of the family $\mathcal F$ then $O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$ is an open neighborhood of the point $x$ disjoint from $F$.

This shows that every point $x\in X\setminus F$ has an open neighborhood $U_x\subseteq X\setminus F$. (More specifically, $U_x=O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$.)

Consequently, $X\setminus F$ is open, since we get $X\setminus F=\bigcup\limits_{x\in X} U_x$, i.e., it is a union of open sets.

score 1 · Accepted Answer · answered May 29 '19 at 08:08

1

Note it suffices to prove that $\overline{\bigcup_{\alpha}F_{\alpha}} = \bigcup_{\alpha}\overline{F}_{\alpha}$, as each $F_{\alpha}$ is closed.

The inclusion $\bigcup_{\alpha}\overline{F}_{\alpha} \subset \overline{\bigcup_{\alpha}F_{\alpha}}$ is clear.

Conversely, if $x \in \overline{\bigcup_{\alpha}F_{\alpha}}$ and not in $\bigcup_{\alpha}\overline{F}_{\alpha}$, take $V$ to be an open neighbourhood of $x$ that touches only finitely many $F_{\alpha}$, say $F_1,...,F_n$.

Note $x \notin F_{\alpha}$ for any $\alpha$ so $x \in \bigcap_{i=1}^n X - F_i$ and therefore in $V \cap \bigcap_{i=1}^n X - F_i$ which is open and touches none of the $F_{\alpha}$ - contradiction.

answered May 29 '19 at 08:08

Mariah

3,760

why $x\not\in F_\alpha$ for any $\alpha$ ? – May 29 '19 at 20:02
By assumption we had $x \notin \bigcup_{\alpha}\overline{F_{\alpha}}$. – Mariah May 29 '19 at 20:25
ah :) right. I have other doubt, why $V \cap \bigcap_{i=1}^n X - F_i$ touches none of the $F_\alpha$ ? I can see that $\bigcap_{i=1}^n X - F_i$ does touch none $F_\alpha$ but cannot see $V$ doesn't touch $F_\alpha$. Before you said it touches finitely many – May 29 '19 at 20:29
@Isa so we have that $V$ touches only $F_1,...,F_n$. That means any subset of $V$ can touch only $F_1,...,F_n$ (no more than those). The intersection $\cap_{i = 1}^n X - F_i$ is disjoint from ALL $F_i$. So if we intersect it with $V$ how can this subset of $V$ intersect any $F_\alpha$? – Mariah May 29 '19 at 20:44
yes, $V$ intersect $F_1,...,F_n$ thus cannot intersect any $F_\alpha$ but in the last line of your answer you said 'touches none of the $F_\alpha$, why did you say none? – May 29 '19 at 20:53
The open set $V \cap \bigcap_{i=1}^n X - F_i$ can not touch any $F_{\alpha}$. This is because $V$ touches at most $F_1,...,F_n$, and the intersection $\bigcap_{i=1}^n X - F_i$ is disjoint from all $F_1,...,F_n$. – Mariah May 29 '19 at 20:57
why is that the contradiction? – May 29 '19 at 21:06
@Isa go over the argument. We assumed $x \in \overline{\bigcup_{\alpha}F_{\alpha}}$. So any neighbourhood of $x$ has to touch the union. The neighbourhood $V \cap \bigcap_{i=1}^n X - F_i$ cannot. – Mariah May 29 '19 at 21:19
got it, thank you Mariah. Can I substitute this $\vert{\alpha\in I:F_\alpha\cap V\neq\emptyset}\vert<\aleph_0$ with to be an open neighbourhood of that touches only finitely many $$, say $_1,...,.$ without problems or a proof is required? – May 29 '19 at 21:25
@Isa no, it's pretty obvious that it means the same thing. – Mariah May 29 '19 at 21:31
oh ok. I'm fine then:) since the $\aleph_0$ looks kinda scary to work with. – May 29 '19 at 21:34
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@Isa as long as you realise that $n$ depends on the $x$ though. Every point has an open neighbourhood that intersects finitely many (that is what $< \aleph_0$ means!) $F_\alpha$. How many exactly depends on the $x$ we're looking at, could be $0$ could be $491$. – Henno Brandsma May 29 '19 at 22:48
@HennoBrandsma that's a good point. – Mariah May 29 '19 at 23:15
@HennoBrandsma oh ok, thank you.The thing is that in none of the exercises before I've seen $\aleph_0,$ and I thought one has to work with cardinality or the definition of $\aleph_0$ which has to do with set theory which I know absolutely nothing about it. – May 30 '19 at 17:12
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@Isa $\aleph_0$ is the first infinite cardinal, by definition. So if $n$ is strictly smaller, it’s not an infinite cardinal so a finite cardinal ( a natural number). That’s all you need for this statement. – Henno Brandsma May 30 '19 at 17:15
@HennoBrandsma ah. Not difficult then. :-) – May 30 '19 at 17:37
@Isa the author probably thought it was "cool" to have $\aleph_0$ in the statement.. $\beth$ and $\daleth$ are also used in set theory. Maybe some other branch of maths should adopt some Arabic letters as well? – Henno Brandsma May 30 '19 at 17:37
@HennoBrandsma no, I think this is already sometimes confusing like to introduce weirder symbols. Of course that if one had all the free time in the world, then it would be really nice to try to understand proofs that their main point is to confuse you. – May 30 '19 at 17:46

William Elliot · Answer 3 · 2019-05-29T22:37:44.740

As the empty set is in V(X), there is a counter example.

Let F$_j$ = [-j,j] for all j in (0,1) and V be the empty set.
Though V intersects the F's finite (zero) many times,
$\cup${ F$_j$ : j in (0,1) } = (-1,1) is not closed.

That "for all x in X" is superfluous as x is not ever used.
Perhaps you were asking about locally finite collections.
Collections with for all x, exists open set V with x in V and V intersecting the F's finitely many times.
Using V(X) instead of the topology is useless complexity.

$\vert\{\alpha\in I:F_\alpha\cap V\neq\emptyset\}\vert<\aleph_0\implies \bigcup\limits_{\alpha\in I} F_\alpha$ is closed.

3 Answers3