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In the beginning of chapter 4 of Dr. Pete Clark's convergence notes:

http://math.uga.edu/~pete/convergence.pdf

Theorem 4.1 (page 13) asserts the equivalence of 5 conditions. After making a nice observation involving De Morgan's law and applying a previous proposition, the problem of proof is reduced to showing the equivalence of the following two statements (extracted from Theorem 4.1 referenced above):

Let $X$ be a topological space.

(b) Every net in $X$ has a limit point.

(e) For every family $\{F_{\alpha}\}_{\alpha\in J}$ of closed subsets with the finite intersection property (the finite intersection of any finite subcollection is non-empty), $\cap_{\alpha\in J}F_{\alpha}\neq \phi$.

For the $(b)\Rightarrow (e)$ direction, it is stated that such a family $F:=\{F_{\alpha}\}_{\alpha\in J}$ is directed under reverse set inclusion (this is used to create a net to which (b) is applied).

I cannot see this.

Of course it is a partially ordered set. But I do not think it needs to be directed. For example, if $X = [0,1]$, $F_{0} = [0,\frac{1}{2}]$, and $F_{1} = [\frac{1}{2},1]$ then $F:=\{F_{0},F_{1}\}$ satisfies the finite intersection property but is not directed. That is, there is no $A\in F$ such that $A\subset F_{0}$ and $A\subset F_{1}$.

Is there a way to argue away this type of case so that we can safely assume that we have a directed set?

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After some thought, I realize that this example immediately satisfies the conclusion of (e), so in this case there is no need to use (b) at all. Maybe I can use this idea to argue away any cases where $F$ is not directed. –  Kyle Schlitt May 12 '12 at 23:24

1 Answer 1

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There's a slight oversight. A collection of closed sets with the finite intersection property isn't necessarily directed by reverse inclusion, but it can be extended to a family that is so directed.

If $\mathscr{F}$ is a family of closed sets with the finite intersection property, let $$\overline{\mathscr{F}}=\left\{\bigcap\mathscr{F}_0:\mathscr{F}_0\text{ is a finite subset of }\mathscr{F}\right\}\;.$$ In other words, $\overline{\mathscr{F}}$ is $\mathscr{F}$ together with all intersections of finite subfamilies of $\mathscr{F}$. In technical terminology, $\overline{\mathscr{F}}$ is the closure of $\mathscr{F}$ under (taking) finite intersections. It's easy to check that $\overline{\mathscr{F}}$ also has the finite intersection property, and $\overline{\mathscr{F}}$ is directed under $\supseteq$. As you'll see, this extension is harmless, because $\overline{\mathscr{F}}$ doesn't contain any set that wasn't in some sense implicitly there in $\mathscr{F}$.

Now for each $F\in\overline{\mathscr{F}}$ choose $x_F\in F$; the assigment $F\mapsto x_F$ is a net $\mathbf x$ in $X$. Let $x$ be a limit point of $\mathbf x$, and suppose that there is some $F\in\overline{\mathscr{F}}$ such that $x\notin F$. Let $U=X\setminus F$; $U$ is open, and $x\in U$, so there is some $G\in\overline{\mathscr{F}}$ such that $G\subseteq F$ and $x_G\in U$. But then $$x_G\in U\cap G\subseteq U\cap F=(X\setminus F)\cap F=\varnothing\;,$$ which is a wee bit difficult, to say the least! Hence $x\in F$ for all $F\in\overline{\mathscr{F}}\supseteq\mathscr{F}$, and thus $x\in\bigcap\mathscr{F}$.

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This resembles (to me) the technique of passing to a subsequence which is commonly done without comment. This will be a good one to remember. As usual, many thanks to you sir! –  Kyle Schlitt May 13 '12 at 1:01

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