Definition Let $(S,\leq)$ is a complete lattice iff $\forall T \subset S :$T admits both a supremum and an infimum.


  1. The collection of all filters on X contained in a given ultrafilter is a complete lattice with 0 and 1
  2. If a family of filters has a supremum then the filters of the family are all contained in some single ultrafilter.

Attempt for:

  1. Let $U$ an ultrafilter, $F=\{\mathbb{F}\subset X | \mathbb{F}$ is a filter and $ \mathbb{F}\subset U\}$. Let $T \subset F$ then $T$ is a filter and $T \subset U$, but I don't see how to prove that 0 is the infimum and 1 is the supremem, it is what I understand for this exercise.
  2. Let $\{\mathbb{F}_{\alpha}\}_{\alpha \in I}$ the family of filters such that exists supremum $\mathbb{G}$, i.e., $\mathbb{F_{\alpha}}\subset \mathbb{G}$ $\forall \alpha \in I$. Let $\mathbb{F}_{\alpha_{i}}$,$\mathbb{F}_{\alpha_j}$, by theorem exists $U_{\alpha_i}, U_{\alpha_j}$ ultrafilters, but I need to prove that $U_{\alpha_i} \neq U_{\alpha_j}$ but I don't see how.

Could you guide me to the right procedure?

  • $\begingroup$ In the first exercise you must decide which members of $F$ are the $0$ and $1$ of the lattice, i.e., the minimum and maximum elements. What is the largest filter belonging to $F$? That filter is the $1$ of $F$. What is the smallest filter belonging to $F$? That is the $0$ of $F$. You also have to show that every $A\subseteq F$ has both a supremum and an infimum in $F$. In the second exercise I suggest assuming that there are $\alpha,\beta\in I$ such that $U_\alpha\ne U_\beta$ and showing that in that case $\Bbb F_\alpha$ and $\Bbb F_\beta$ have no supremum. $\endgroup$ Apr 19, 2020 at 19:27

1 Answer 1


As to 1.:

If we have the collection of all filters contained in $\mathcal{U}$, an ultrafilter on $X$, the $0$ of the lattice is the filter $\{X\}$ and the $1$ is $\mathcal{U}$ itself.

If $\mathcal{F}_i, i \in I$ is a set of such filters, its sup is the filter generated by $\bigcup_i \mathcal{F}_i$ (a filter base), while the inf is $\bigcap_i \mathcal{F}_i$, which is a filter directly.

And for 2: I’d $\mathcal{F}_i, i \in I$ is a set of filters with supremum the filter $\mathcal{G}$, this implies that all these filters are a subset of any ultrafilter $\mathcal{U}$ we choose (Zorn/AC) that extends that $\mathcal{G}$.

  • $\begingroup$ What if $\mathscr{U}$ is fixed? $\endgroup$ Apr 19, 2020 at 19:45
  • $\begingroup$ @BrianM.Scott The $\mathcal{U}$ is fixed for this lattice argument. That’s what 1 intends $\endgroup$ Apr 19, 2020 at 19:48
  • 1
    $\begingroup$ No, I mean what if $\mathscr{U}$ is a principal ultrafilter? It doesn’t even contain the cofinite filter. Isn’t $\{X\}$ the $0$ no matter what $\mathscr{U}$ is? $\endgroup$ Apr 19, 2020 at 19:49
  • $\begingroup$ @BrianM.Scott it seems so, yes. In my head the cofinite filter is minimal. $\endgroup$ Apr 19, 2020 at 21:07
  • $\begingroup$ Me too: when I first looked at it I was sure that I’d have to separate the fixed and free cases! $\endgroup$ Apr 19, 2020 at 21:11

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