A maximal element in the partially ordered set (for the inclusion relation) of all filters on a set $X$ is called an ultrafilter on $X$


If $\mathcal{F}$ is a filter on a set $X$, then there exists an ultrafilter $\mathcal{U}$ such that $\mathcal{F} \subseteq \mathcal{U}$

Proof my book provides:

Consider the set $\mathcal{X} := \{\mathcal{G} \mid \mathcal{G}\mathrm{\ filter}, \mathcal{F} \subseteq \mathcal{G}\}$. Then $(\mathcal{X}, \subseteq)$ is a poset and it is inductive ordened. Hence, the existence of the ultrafilter $\mathcal{U}$ follows by Zorn's lemma. $\square$

My questions:

1) I already proved that the set of all filters on $X$ is partially ordered. Does this imply that the poset ($\mathcal{X}, \subseteq)$ is inductive ordened as well? I think not, but my book claims it does.

2) The proof in my book gives a maximal element of the set $\mathcal{X}$, which is a subset of the set of all filters. Hence, is the maximal element found by Zorn's lemma also a maximal element in the set of all filters on $X$? This should be true (but I can't prove it), in order for $\mathcal{U}$ (that is found in the proof) to be an ultrafilter by definition.


I don't know what is the definition of inductive order in that book, but in Wolfram one possibility is a partially ordered set in which every totally ordered subset has an upper bound.

This is the case of $(\mathcal X,\subseteq)$ because if $(\mathcal G_i)_{i \in I}$ is a chain of filters containing $\mathcal F$, then $\mathcal G=\bigcup_{i\in I}\mathcal G_i$ is an upper bound of that chain.
So you are in conditions of applying Zorn's Lemma, and conclude that $(\mathcal X,\subseteq)$ has a maximal element which is, therefore, an ultrafilter.

  • $\begingroup$ Yes, it is that definition. But why is a maximal element of $(\mathcal{X}, \subseteq)$ an ultrafilter? An ultrafilter must be a maximal element of the set of all filters. $\endgroup$ – user370967 Mar 10 '18 at 18:33
  • $\begingroup$ @Math_QED A maximal filter among those that contain $F$ is a maximal filter overall. $\endgroup$ – amrsa Mar 10 '18 at 18:37
  • $\begingroup$ Yes, ok. I see it. And for the second question, is the set inductive ordened because the set of all filters on $X$ is inductive ordened? I can't see how this would follow, but I can see it is true by doing the same proof. $\endgroup$ – user370967 Mar 10 '18 at 18:46
  • $\begingroup$ @Math_QED if $\mathcal H$ is a maximal element (it can have more than one) then the implication you're asking follows from $\mathcal H$ being maximal. All elements is $\mathcal X$ contain $\mathcal F$. $\endgroup$ – amrsa Mar 10 '18 at 18:46
  • $\begingroup$ @Math_QED It is obvious that a maximal element in $\mathcal X$ is maximal in the set of all filters. For if $H$ is maximal in $\mathcal X$, and $H \subseteq K$, where $K$ is another filter, then $F \subseteq K$ (because $F \subseteq H$) and thus $K \in \mathcal X$, yielding $K=H$ because $H$ is maximal in $\mathcal X$. $\endgroup$ – amrsa Mar 10 '18 at 18:50

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