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Homeproof that every filter is contained in an ultrafilter

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# proof that every filter is contained in an ultrafilter

Let $Y$ be the set of all non-empty subsets of $X$ which are not contained in $\mathcal{F}$. By Zermelo’s well-orderding theorem, there exists a relation ‘$\succ$’ which well-orders $Y$. Define $Y^{{\prime}}=\{0\}\cup Y$ and extend the relation ‘$\succ$’ to $Y^{{\prime}}$ by decreeing that $0\prec y$ for all $y\in Y$.

We shall construct a family of filters $S_{i}$ indexed by $Y^{{\prime}}$ using transfinite induction. First, set $S_{0}=\mathcal{F}$. Next, suppose that, for some $j\in Y$, $S_{i}$ has already been defined when $i\prec j$. Consider the set $\bigcup_{{i\prec j}}S_{i}$; if $A$ and $B$ are elements of this set, there must exist an $i\prec j$ such that $A\in S_{i}$ and $B\in S_{i}$; hence, $A\cap B$ cannot be empty. If, for some $i\prec j$ there exists an element $f\in S_{i}$ such that $f\cap j$ is empty, let $S_{j}$ be the filter generated by the filter subbasis $\bigcup_{{i\prec j}}S_{i}$. Otherwise $\{j\}\cup\bigcup_{{i\prec j}}S_{i}$ is a filter subbasis; let $S_{j}$ be the filter it generates.

Note that, by this definition, whenever $i\prec j$, it follows that $S_{i}\subseteq S_{j}$; in particular, for all $i\in Y^{{\prime}}$ we have $\mathcal{F}\subseteq S_{i}$. Let $\mathcal{U}=\bigcup_{{i\prec j}}S_{i}$. It is clear that $\emptyset\notin\mathcal{F}$ and that $\mathcal{F}\subseteq\mathcal{U}$.

It is easy to see that $\mathcal{U}$ is a filter. Suppose that $A\cap B\in\mathcal{U}$. Then there must exist an $i\in Y^{{\prime}}$ such that $A\cap B\in S_{i}$. Since $S_{i}$ is a filter, $A\in S_{i}$ and $B\in S_{i}$, hence $A\in\mathcal{U}$ and $B\in\mathcal{U}$. Conversely, if $A\in\mathcal{U}$ and $B\in\mathcal{U}$, then there exists an $i\in Y^{{\prime}}$ such that $A\in S_{i}$ and $B\in S_{i}$. Since $S_{i}$ is a filter, $A\cap B\in S_{i}$, hence $A\cap B\in\mathcal{U}$. By the alternative characterization of a filter, $\mathcal{U}$ is a filter.

Moreover, $\mathcal{U}$ is an ultrafilter. Suppose that $A\in\mathcal{U}$ and $B\in\mathcal{U}$ are disjoint and $A\cup B=X$. If either $A\in\mathcal{F}$ or $B\in\mathcal{F}$, then either $A\in\mathcal{U}$ or $B\in\mathcal{U}$ because $\mathcal{F}\subset\mathcal{U}$. If $A\in Y$ and $A\in S_{A}$, then $A\in\mathcal{U}$ because $S_{A}\subset\mathcal{U}$. If $A\in Y$ and $A\notin S_{A}$, there must exist $x\in\mathcal{U}$ such that $A\cap x$ is empty. Because $B$ is the complement of $A$, this means that $x\subset B$ and, hence $B\in\mathcal{U}$.

This completes the proof that $\mathcal{U}$ is an ultrafilter — we have shown that $\mathcal{U}$ meets the criteria given in the alternative characterization of ultrafilters.

## Mathematics Subject Classification

54A20*no label found*

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## Comments

## err...

I have never heard of a "subset of X" which is "not contained in X"... this is a very, very careless mistake.

## Re: err...

Evidently you're new to the site. Ray (user rspuzio) is one of the PlanetMath's most active contributors; please try not to be so careless yourself and properly follow the site's protocol by filing your correction via the "correct" button at the bottom of the object.

## Re: err...

I can do this, but the proof I know is fairly different than this one.

## Re: err...

I am now reading the guidelines, sorry for that...

## Re: err...

It was a simple misprint, which is now fixed.

## Re: err...

In that case, please add your proof. There are certainly different

ways to go about proving the result and having more proofs of the

result presented will improve the coverage of the subject.