# Proof of Kelley's Theorem

So, still working through the convergence notes here:

http://math.uga.edu/~pete/convergence.pdf

And I was wondering if somebody could help me finish the proof of Kelley's Theorem, which states that every net admits a universal subnet.

It's on page 12, and I followed up to the end of the second paragraph, but then when we assume that there is at least one $Y\in \mathcal A$ such that the net $\bf x$ is not cofinal in $Y\cap A$, I am lost after that.

What I do understand is: why we are able to assume this (since the negation of that assumption leads to the desired result as proved in the previous paragraph)

What I am confused about: I am pretty sure the $Z$ is a typo as it is not anywhere else, and the obvious guess is that it should be a $A$ (or maybe there is a missing quantifier and it really should be a $Z$?), but even with this modification, I still confused how the previous case applies since it seemed to be a much different hypothesis.

That is, (assuming my modification is correct, which it may very well not be) we have the assumption that $x$ is eventually in $X\backslash (A\cap Y)$ for a particular $Y$, but in the previous case we had a much stronger assumption for the argument.

I realize most of this makes no sense unless one has read the proof in the notes linked above, and this is a lot of me to ask of anyone, but I spent quite a bit of time on this and I am definitely getting nowhere.

I found a few other proofs of this theorem, but most of them seem to utilize the very machinery that these notes are working towards developing (filters), so they are not appropriate to use as as helper here.

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First, the $Z$ is indeed a typo for $A$. Secondly, there does seem to be a bit of an omission in the argument in the third paragraph.
We have a $Y_0\in\mathcal A$ such that $\mathbf x$ is not cofinal in $A\cap Y_0$ and hence is eventually in $X\setminus(A\cap Y_0)$. For convenience let $Z=X\setminus(A\cap Y_0)$. For any $Y\in\mathcal{A}$, $\mathbf x$ is eventually in $Z$ and cofinally in $Y$, so $\mathbf x$ is cofinally in $Y\cap Z$. Now use the argument of the previous paragraph: the family $\mathcal{A}'$ of sets containing $Y\cap Z$ for some $Y\in\mathcal{A}$ satisfies (i)-(iii), so $\mathcal{A}'=\mathcal{A}$ by the choice of $\mathcal{A}$, and hence $Z\in\mathcal{A}$. But in particular we have $Y_0\cap Z\in\mathcal{A}$, and $$Y_0\cap Z=Y_0\cap\Big(X\setminus(A\cap Y_0)\Big)=Y_0\setminus(A\cap Y_0)=Y_0\setminus A\;,$$ and it now follows from (ii) that $X\setminus A\in\mathcal{A}$.
(Speaking of filters, you probably noticed that families satisfying (i) and (ii) are precisely filters on $X$.)