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I'm trying to prove the following properties:

Given a boolean algebra $B$,

  • $U\subseteq St(B)$ is open if and only if $\{c\in B : N_c\subseteq U\}$ is an ideal on $B$;
  • $F\subseteq St(B)$ is closed if and only if $\{c\in B : N_c\supseteq F\}$ is a filter on $B$.

I specify that we are considering $(St(B),\tau_B)$, where $St(B)=\{G\subseteq B : G \text{ is an ultrafilter}\}$, and $\tau_B$ is the topology generated by $\{N_b=\{G\in St(B): b\in G\}:b\in B\}$. The clopen sets of $\tau_B$ are the $N_b$ for $b\in B$, and form a basis for $\tau_B$.

I proved the $\Rightarrow$ implications, but actually without using the hypothesis that $U$ and $F$ are respectively open and closed, in an elementary way. I am stuck with the $\Leftarrow$ implications. Here is my attempt:

Let $I=\{c\in B : N_c\subseteq U\}$; $I$ is closed under $\vee$ because if $c,d\in I$ then $N_c,N_d\subseteq U$, and $N_c\cup N_d =N_{c\vee d}\subseteq U$.

Then let $c\in I$ and $b\leq c$. Therefore $N_c\subseteq U$, and if $G\in N_b$ we have that $c\in G$, so $G\in N_c$. We conclude that $N_b\subseteq N_c\subseteq U$.

The proof of $\{c\in B : N_c\supseteq F\}$ being a filter is almost the same. What's wrong with it? Any suggestions about the other implications? Thank you in advance.

mimar
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    Why are you trying to prove these statements? They are not true... – Eric Wofsey Sep 05 '21 at 19:34
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    Perhaps what was intended is this: There is a bijection between open sets in $\mathrm{St}(B)$ and ideals in $B$, given by $U\mapsto {c\in B\mid N_c\subseteq U}$ for $U$ open and $I\mapsto \bigcup_{c\in I} N_c$ for $I$ an ideal. And similarly for closed sets and filters. – Alex Kruckman Sep 05 '21 at 19:59
  • I found the statements on some notes, written as I wrote in the question. I am sure that the meaning of those statements is actually what @Alex Kruckman said, this makes totally sense now! Thank you. – mimar Sep 06 '21 at 14:07

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