Skip to content
January 2, 2018 / porton

Very easy solution of my old conjecture

Like a complete idiot, this took me a few years to disprove my conjecture, despite the proof is quite trivial.

Here is the complete solution:

Example [S]\ne\{\bigsqcup^{\mathfrak{A}}X \mid X\in\mathscr{P} S\}, where [S] is the complete lattice generated by a strong partition S of filter on a set.

Proof Consider any infinite set U and its strong partition \{\uparrow^U\{x\} \mid x\in U\}.

\{\bigsqcup^{\mathfrak{A}}X \mid X\in\mathscr{P} S\} consists only of principal filters.

But [S] obviously contains some nonprincipal filters.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: