I'm working through problems in Bredon's Topology and Geometry, and I've gotten stuck on Chapter I Section 8 Problem 8(b), so I thought I'd give this site a try. The problem goes as follows:
"Let $A$ be an uncountable set. For each $\alpha \in A$ let $X_{\alpha} = \{0, 1\}$ with the discrete topology. Put $X = \times_{\alpha \in A} X_{\alpha}$. That is, $X = \{0, 1\}^{A}$. Let $p \in X$ be the point with all components $p_{\alpha} = 1$. Show that there is no neighborhood basis for $p$ simply ordered (i.e. linearly ordered) by inclusion."
The rest of the problem is straightforward, but this part seems to require some set theory. Something like this: Suppose $A$ is an uncountable set, $S$ the collection of all finite subsets of $A$, and $T$ is a subcollection of $S$ linearly ordered by inclusion. Then there must be some $s \in S$ such that $s$ is not contained in $t$ for any $t \in T$. Can anyone see how to prove this instead?
EDIT: I've realized that my "reformulation" in terms of set theory isn't exactly the same problem, because it assumes that given a neighborhood basis for $p$ that is simply ordered by inclusion, you can find a simply ordered neighborhood basis consisting only of basic open sets, which may not always be possible. If anyone can see how to prove that it is always possible in this specific case, I greatly would appreciate it.