Some problems on locally finite sets

Question

This is a series of problems on locally finite sets with a few I am stuck on. Can the community provide some help on how to proceed?

Definition. Let $(X,\tau)$ be topological space. A set $\mathcal S$ of subsets of $X$ is said to be locally finite in $(X,\tau)$ if each point $x \in X$ has a neighborhood $N_x$ such that $N_x \cap S= \varnothing$ for all but a finite number of $S \in \mathcal S$.

Prove the following statements:

$(\rm i)$ If the set $\mathcal S$ of subsets of $X$ is finite, then $\mathcal S$ is locally finite.

This is trivial.

$(\rm ii)$ If the set $\mathcal S$ is such that every point of $X$ lies in at most one $S \in \mathcal S$, then $\mathcal S$ is locally finite.

I am not sure if this statement is true. Let us assume that $\mathcal S$ is infinite, otherwise, from part $(\rm i)$, it is locally finite. Now $X$ must also contain an infinite number of elements. If $T$ is the indiscrete topology, where the only open sets are the whole set and the empty set, then even if every point of $X$ lies in one $S$, $\mathcal S$ is not locally finite because the whole space is the only open neighborhood of each point, and its intersection with any $S$ is non-empty because they are all part of the whole space, so I think $(\rm ii)$ is not a true statement.

$(\rm iii)$ Let $X$ be an infinite set and $\tau$ the finite-closed topology on $X$. If $\mathcal S$ is the set of all open sets in $(X,\tau)$, then $\mathcal S$ is not locally finite.

Every open set in the cofinite topology has a non empty intersection with every other open set so $\mathcal S$ cannot be locally finite.

$(\rm iv)$ Let $\mathcal S$ be a locally finite. Define $\mathcal T$ to be the set of all closed sets $T=\operatorname{cl}(S)$ for $S \in \mathcal S$. Then $\mathcal T$ is locally finite.

So, this is saying that the closure of each $S \in \mathcal S$ is still locally finite. Is this a proof by contradiction or is a direct proof possible? If a open neighborhood of a point has an empty intersection with all but finite number of $S$, is there a reason to believe this would change if we took the closure of all $S$?

$(\rm v)$ If $\mathcal S$ is an infinite set of subsets of a infinite set $X$, and $(X,\tau)$ is a compact space, then $\mathcal S$ is not locally finite.

Suppose $\mathcal S$ is locally finite. Then for each $x \in X$, there exist a $U_i$ such that $U_i$ has an empty intersection with all but finite number of $S \in \mathcal S$. The set $U_i$ covers $X$, so there exist a finite subcover. How can I show a contradiction from here?

$(\rm vi)$ Let $\mathcal S$ be an uncountable set of subsets of $X$. If $\mathcal S$ is a cover of the space $(X,\tau)$ and $(X,\tau)$ is either a Lindelöf space or a second countable space, then $\mathcal S$ is not locally finite.

Let us assume $X$ is second countable first and also assume $\mathcal S$ is locally finite. Then for each $x \in X$, there exist a $U_i$ such that $U_i$ has an empty intersection with all but finite number of $S \in \mathcal S$. How should I proceed to show a contradiction?

Answer 1

Yes, (ii) is false. There are even counterexamples in $\Bbb R^2$. For $n\in\Bbb N$ let

$$S_n=\left\{\langle x,y\rangle\in\Bbb R^2:x>0\text{ and }y=nx\right\}\,,$$

and let $\mathscr{S}=\{S_n:n\in\Bbb N\}$; then $\mathscr{S}$ is pairwise disjoint, but every open nbhd of the origin meets every $S_n$, so $\mathscr{S}$ is not locally finite at the origin.

For (iv) just note that if $U$ is open, and $U\cap S=\varnothing$, then $X\setminus U$ is a closed set containing $S$, so $X\setminus U\supseteq\operatorname{cl}(S)$, and therefore $U\cap\operatorname{cl}(S)=\varnothing$.

You’ve done most of the work for (v). If $\mathscr{S}$ is locally finite, each $x\in X$ has an open nbhd $U_x$ that meets only finitely many members of $\mathscr{S}$; let $\mathscr{U}=\{U_x:x\in X\}$. $X$ is compact, so there is a finite $F\subseteq X$ such that $\mathscr{U}_0=\{U_x:x\in F\}$ covers $X$. For each $x\in F$ let $\mathscr{S}_x=\{S\in\mathscr{S}:U_x\cap S\ne\varnothing\}$, and let $\mathscr{S}_F=\bigcup_{x\in F}\mathscr{S}_x$. Each $\mathscr{S}_x$ is finite, so $\mathscr{S}_F$ is finite. But every member of $S$ meets at least one member of $\mathscr{U}_0$, so $\mathscr{S}=\mathscr{S}_F$, and $\mathscr{S}$ is therefore finite.

Finally, (vi) is really the same argument. First, though, note that every second countable space is automatically Lindelöf, so it suffices to prove (vi) for Lindelöf spaces. Just imitate the preceding argument: this time you get a countable $F\subseteq X$ such that $\mathscr{U}_0$ covers $X$. Define $\mathscr{S}_x$ for $x\in F$ as before, and likewise $\mathscr{S}_F$. Then $\mathscr{S}_F$ is the union of countably many finite sets, so it is countable, and $\mathscr{S}=\mathscr{S}_F$ just as before.