# Moore–Smith convergence; Universal nets

I am looking at John Kelley's General Topology. This is exercise J from chapter $2$. I seem to be having trouble with problems c) and d).

c) Prove the following

Lemma: If $S$ is a net in $X$, then there exists a family $C$ of subsets of $X$ such that:

1) $S$ falls often in each element of $C$.

2) The intersection of any two elements of $C$ stays in $C$.

3) $(\forall A\subset X)(A\in C \lor X\setminus A\in C)$

HINT: Show that if the family C is maximal regarding the first two properties, then the third one holds. (I believe maximal means maximal regarding set theoretic inclusion.)

d)

Prove that every net has a universal subnet.

A side question is why if $f:X \to Y$ and $S$ is a universal net in X then $f\circ S$ is a universal net in $Y$. Some may deem it trivial but I could not quite get convinced.

p.s. I apologize for any confusion of terminology as I am translating my Bulgarian copy of the textbook, and I don't have any mathematical dictionaries. Corrections are welcome.

Edit: The book advices to use theorem 2.5 to prove d).

Lemma 2.5: Let $S$ be a net and $A$ a family of sets such that the intersection of any two elements of $A$ contains and elements of $A$, and $S$ is often in each element of $A$. Then there is a subnet of $S$ that is almost completely in each element of $A$.

Let $\langle D,\le\rangle$ be the directed set on which $S$ is based. For each $d\in D$ let $T_d=\{S(e):d\le e\in D\}$, and let $\mathscr{C_0}=\{C\subseteq X:C\supseteq T_d\text{ for some }d\in D\}$. In words, $T_d$ is the tail of $S$ determined by the element $d\in D$, and $\mathscr{C_0}$ is the family of subsets of $X$ that contain some tail of $S$.

1. Let $C\in\mathscr{C_0}$ and $d\in D$. There is an $e\in D$ such that $C\supseteq T_e$, and there is an $e'\in D$ such that $d,e\le e'$, and clearly $S(e')\in T_d\subseteq C$. Thus, $S$ is frequently in $C$.

2. A similar argument shows that $\mathscr{C_0}$ is closed under finite intersections. If $C_0,C_1\in\mathscr{C_0}$, there are $d_0,d_1\in D$ such that $C_i\supseteq T_{d_i}$ for $i=0,1$. Choose $d\in D$ so that $d_0,d_1\le d$; then $T_d\subseteq T_{d_0}\cap T_{d_1}\subseteq C_0\cap C_1$, so $C_0\cap C_1\in\mathscr{C_0}$.

Now let $\mathscr{C}$ be a maximal (with respect to $\subseteq$) subset of $\wp(X)$ containing $\mathscr{C}_0$ and satisfying (1) and (2). You can use Zorn’s lemma or transfinite recursion to get $\mathscr{C}$; I can add either of these arguments if you like.

Let $A\subseteq X$ be arbitrary, and suppose that $A\cap C\ne\varnothing$ for all $C\in\mathscr{C}$. Let $\mathscr{F}=\langle\mathscr{C}\cup\{A\}\rangle$, the filter generated by $\mathscr{C}\cup\{A\}$; clearly $\mathscr{F}$ satisfies (2). If $\mathscr{F}$ does not satisfy (1), it must be because $T_d\cap C\cap A=\varnothing$ for some $d\in D$ and $C\in\mathscr{C}$. But $T_d\cap C\in\mathscr{C}$, so this is impossible, and therefore $\mathscr{F}$ satisfies (1) as well as (2). It follows from the maximality of $\mathscr{C}$ that $\mathscr{F}=\mathscr{C}$ and hence that $A\in\mathscr{C}$.

Corrected paragraph: Now suppose that $A\cap C_0=\varnothing$ for some $C_0\in\mathscr{C}$, and let $B=X\setminus A$. Let $C\in\mathscr{C}$ be arbitrary, and let $C_1=C_0\cap C\in\mathscr{C}$. Then $A\cap C_1=\varnothing$, and $S$ is frequently in $C_1$, so $S$ is frequently in $C_1\cap B\subseteq C\cap B$. In particular $C\cap B\ne\varnothing$ for all $C\in\mathscr{C}$, and by the previous paragraph we must have $B\in\mathscr{C}$. Thus, $\mathscr{C}$ satisfies (3) as well as (1) and (2).

For (d), apply Lemma (not Theorem) $2.5$ of your edit to the family $\mathscr{C}$ to get a subset $S'$ of $S$ that is eventually in each $C\in\mathscr{C}$. Then by (3) you know that for any $A\subseteq X$, $S'$ is eventually in $A$ or eventually in $X\setminus A$, so $S'$ is universal.

Finally, let $f:X\to Y$ be arbitrary, and let $S:D\to X$ be a universal net. Let $A\subseteq Y$ be arbitrary. Then $S$ is eventually in either $f^{-1}[A]$ or $X\setminus f^{-1}[A]$. That is, there is a $d\in D$ such that either $T_d\subseteq f^{-1}[A]$, or $T_d\subseteq X\setminus f^{-1}[A]$. In the first case $f[T_d]\subseteq A$, so $f\circ S$ is eventually in $A$. In the second case we use the fact that $X\setminus f^{-1}[A]=f^{-1}[Y\setminus A]$ to argue similarly that $f[T_d]\subseteq Y\setminus A$ and hence that $f\circ S$ is eventually in $Y\setminus A$. Since one of the cases must apply, $f\circ S$ is universal.

• Thank you, Brian. I am familiar with Zorn's lemma. Let me show as exercise, that $\subseteq$ is a partial order and each chain has an upper bound in the desired set. Risking looking stupid I wanted to ask how it follows that each chain has a universal sub-chain now. Given any net can we construct the universal subnet? – superAnnoyingUser Mar 6 '13 at 16:08
• @George: I’m not sure that I understand your question. Once you know that each chain has an upper bound, you know that the partial order has a maximal element, which you then take to be $\mathscr{C}$. What do you mean by a universal subchain of a chain in the partial order? – Brian M. Scott Mar 6 '13 at 16:11
• Yes, precisely, once I know each chain has an upper bound I can use Zorn's lemma. What I meant to ask is how to prove that each net has a universal sub-net, but you have already added that. Excuse my mistake. – superAnnoyingUser Mar 6 '13 at 16:40
• @George: That follows immediately from (c) and Lemma $2.5$: see the second-last paragraph of my answer. – Brian M. Scott Mar 6 '13 at 16:42
• I'm having trouble finding upper bounds for the chains, to show that Zorn's lemma is applicable. Should I argue by admitting the contrary? Where does the contradiction stem from? – superAnnoyingUser Mar 6 '13 at 21:12