STATEMENT: An “ultrafilter” is a filter that is not properly contained in any other filter. Use Zorn’s lemma to show that every filter is contained in an ultrafilter.

PROOF: Let $F$ be the set of all filters that are contained in $X$. Let $\mathscr{C}$ be a chain in F. Then let us take $\mathcal{C}'=\bigcup_{C∈\mathscr{C}}C.$ Then, clearly, $\mathcal{C}'$ is an upper bound for $C$ since given any element, $B$, of our chain we have $ B\subseteq \mathcal{C}'$. Let $A,B∈\mathcal{C}'$. Since $A$ and $B$ are in $\mathcal{C}'$, they must be contained in some filter $\mathcal{F}. $Therefore, $A∩B∈\mathcal{F}$. It therefore follows that $A∩B∈C'$. Furthermore, if $A⊆D$ then since $A∈F$ we have that $D∈F$. So $D∈\mathcal{C}'$. So, $\mathcal{C}'$ is a filter. By Zorn's lemma we conclude that there is a maximal filter(i.e. an ultrafilter) in $F$.

QUESTION: I wanted to see if there were any problems with my proof. First time using Zorn's lemma. I also wanted to see why that if given any ultrafilter in $X$ that if $A\subseteq X$, then either $A$ is in the ultrafilter or $A^c$ is in the ultrafilter, but not both. Also the definition of filter in my case restricts the filter to not have the empty set.

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    $\begingroup$ And your question is...? $\endgroup$ – Asaf Karagila Oct 9 '14 at 17:12
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    $\begingroup$ Your proof is fine. $\endgroup$ – Brian M. Scott Oct 9 '14 at 17:20

First let's do the easier part. Why ultrafilters have the property that either $A$ or its complement are in them: for this you have to prove that if $\cal F$ is a filter, and $A$ is a set which has a non-empty intersection with every element of $\cal F$, then there is a filter $\cal F'$ such that $\cal F\subseteq F'$ and $A\in\cal F'$.

If you prove that part, then from maximality the property ensues.

As for the proof, it's a bit disorganized, and the choice of letters is bad ($F$ is likely to denote a filter, not the partial order, if $C$ is the chain, $C'$ shouldn't denote its union). This will be mitigated if you use different fonts, e.g. $\mathscr F$ for the partial order, and $\mathcal C$ for the chain, and so on.

This causes another problem. You want to show that $C'$ is a filter on $X$. But it's very unclear what you're actually showing there. $B\subseteq C'$, rather than $B\in C'$. Also $C$ is a chain in $F$, so $B\in C$ implies that $B\in F$.

Finally, since you wanted the ultrafilter to extend a fixed filter, you should have concentrated on the filters which extend that fixed filter.

My advice for you is to rewrite it, use more line breaks to clarify the text, use different font styles for different "types" of objects (subsets of $X$ are in usual roman letters; filters in calligraphic font (\mathcal); chains in script \mathscr) or some other notational scheme that will help you be clear about which variables are sets, filters, or sets of filters.

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    $\begingroup$ Thanks for the advice Asaf. I will fix the typographical issues. I'm using texmacs btw, so the copy/paste function doesn't carry over completely error free. $\endgroup$ – Enigma Oct 9 '14 at 18:15

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