# The collection of all filters on X is a complete lattice with 0 and 1 and a family of filters has a supremum

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

Problems

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?

• 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. Apr 19, 2020 at 19:27

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 $$\mathcal{G}$$.

• What if $\mathscr{U}$ is fixed? Apr 19, 2020 at 19:45
• @BrianM.Scott The $\mathcal{U}$ is fixed for this lattice argument. That’s what 1 intends Apr 19, 2020 at 19:48
• 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? Apr 19, 2020 at 19:49
• @BrianM.Scott it seems so, yes. In my head the cofinite filter is minimal. Apr 19, 2020 at 21:07
• Me too: when I first looked at it I was sure that I’d have to separate the fixed and free cases! Apr 19, 2020 at 21:11