# Why is there no neighborhood basis for $(1)$ in $\{0,1\}^{[0, 1]}$ simply ordered by inclusion?

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.

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Your plan seems to be a bit too abstract to have a chance. Which particular topology on $\{0,1\}^A$ you're working with. (For example, if it had been the box product, you would just get the discrete topology (on the same underlying set), in which case singletons are always neighborhood bases). It feels like a good lemma in this case would be to prove that every open set in $X$ must be indifferent to all but countably many coordinates. – Henning Makholm Nov 3 '11 at 1:38
@Henning: I think the intention is to take the product topology, unless some other topology on the product is specified. – Carl Mummert Nov 3 '11 at 2:32
@Carl, I agree with that. The point of my comment was that I couldn't see where the choice of topology entered the proof sketch. Now, I suppose that this is exactly where the "collection of all finite subsets" comes from. – Henning Makholm Nov 3 '11 at 2:36
@Henning: In my mind the key role of the product topology in the problem is in the last sentence of the first paragraph of my answer. – Carl Mummert Nov 3 '11 at 2:39

First let $(a(n) : n \in \mathbb{N} )$ be a countable sequence of distinct points in $A$. For each $n$ let $U_n$ be the basic open set that for all $i \leq n$ restricts coordinate $a(i)$ to be $1$. Then $\mathcal{U} = \{ U_n : n \in \mathbb{N}\}$ is not a neighborhood basis for $p$, because it only restricts a countable number of coordinates. Also the intersection $\mathcal{U}$ contains no open neighborhood of $p$, because the intersection contains no basic open set at all from the product topology.

If there was some linearly ordered neighborhood basis $\mathcal{V}$ of $p$, then we could find a sequence $V_n \in \mathcal{V}$ with $V_n \subseteq U_n$ and $V_{n+1} \subseteq V_n$ for each $n$. Then $(V_n)$ is again not a neighborhood basis for $p$, by a similar argument to the first paragraph. Also, $\bigcap_n V_n$ contains no open set, because it is a subset of $\bigcap_n U_n$, so there is no $V \in \mathcal{V}$ such that $V \subseteq V_n$ for all $n$, hence $\mathcal{V}$ is not a linearly ordered neighborhood basis for $p$, as desired.

The key point is that any linearly ordered neighborhood basis would have to have confinality $\omega$ but at the same time there is no countable neighborhood basis of $p$.

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Ah, so the countable string of $V_n$'s is in some sense "at the end" of the chain, which cannot happen since it would give you a countable neighborhood basis (which clearly can't happen). Thanks for the response! – DA29731 Nov 3 '11 at 5:03

Since the elements of $T$ all have distinct finite cardinalities, $T$ must be countable. It follows that $\cup T$ is also countable, so $A \setminus \cup T$ is nonempty.

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Thanks for the response! This sounds like an important fact about cardinalities I should know: the number of distinct finite cardinalities is countable? Or something like that? – DA29731 Nov 3 '11 at 4:52
If the elements of $T$ have distinct finite cardinalities, then you can make a map $f$ such that $f(n)$ is the element of $T$ with cardinality $n$, if there is such an element. Then $f$ is a possibly partial surjection from $\mathbb{N}$ to $T$, so $T$ is countable. – Carl Mummert Nov 3 '11 at 11:12
That is, $f$ is a bijection from a subset of $\mathbb{N}$ to $T$ (too late to edit my original response). – Carl Mummert Nov 3 '11 at 12:22