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I was reading this discussion of Hindman's Theorem by Leo Goldmakher, and was tripped up by his introduction of a topology on $U(\mathbb N)$. (He is using $U(\mathbb N)$ to denote the space of ultrafilters on the natural numbers.) Midway through page three, Goldmakher says, "There is a natural topology on $U(\mathbb N)$, given by the basis of open sets $\{\mathcal U\in U(\mathbb N):A\in\mathcal U\text{ for some }A\subseteq\mathbb N\}$," and this is all he says about the matter until he uses this topology in the proof of Theorem 3.1 several pages later.

I have two questions about this:

(1): Am I correct in thinking that the topology Goldmakher is defining here is the one generated by the basis $\{\{\mathcal U\in U(\mathbb N):A\in\mathcal U\}:A\subseteq\mathbb N\}$? The set $\{\mathcal U\in U(\mathbb N):A\in\mathcal U\text{ for some }A\subseteq\mathbb N\}$ is (as I am currently interpreting it) just the set $U(\mathbb N)$; but that doesn't make sense in context, and, besides, if he meant $U(\mathbb N)$ that's what he would have written.

(2): I know this is vague (so feel free to ignore it), but are there any useful ways of thinking about this topology that might help me understand it?

A. Howells
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2 Answers2

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The answer to your first question is yes.

For your second question, I like the following. Think of a set $A\subseteq\mathbb{N}$ as a question you're asking about some number $N$ that I know and you're trying to figure out (a la "Twenty Questions" but for large values of twenty) - the question is, "Is $N\in A$?" For instance, you might ask "Is $N$ even?" That is, "Is $N\in \{2, 4, 6, ...\}$?"

Now, you can think of a nonprincipal ultrafilter as a convincing way to cheat. That is, maybe I don't actually have an $N$ in mind - each time you ask a question, I just make something up. That way you never figure out what "$N$" is, since there is no actual $N$!

In order to fool you, I need my answers to be consistent. That is:

  • If you ask "Is $N\in A$?" and I say "Yes," and you then ask "Is $N\in B$?" for some $B\supseteq A$, I'd better say "Yes."

  • If you ask "Is $N\in A$?" and "Is $N\in B$?" and I say "Yes" to each, then I'd better say "Yes" when you ask "Is $N\in A\cap B$?"

  • If you ask "Is $N\in A$?" and I say "No," then I'd better say "Yes" when you ask "Is $N\in \overline{A}$?" (and conversely).

  • If you ask "Is $N\in F$?" for some finite set $F$, I'd better say "no", since otherwise you'll be able to pin me down to a single number (I'm trying to cheat, remember?).

So you can think of an arbitrary ultrafilter as a strategy - that is, a way for me to play this game, without every obviously lying. Sometimes (principal ultrafilters) I'm not cheating, while other times (nonprincipal ultrafilters) I am. In this context, a basic open set corresponds to a question: the set of ultrafilters containing $A$ is, essentially, the set of strategies which make me answer "Yes" when you ask "Is $N\in A$?"

Note the twist here: "Is $N$ in $A$?" is the same question as "Is $A$ in $\mathcal{U}$?". An ultrafilter is basically a "pseudo-number": it behaves like a natural number in the context of the game above. Specifically, we can identify a number with the set of questions about it whose answer is "yes" - this is just the principal ultrafilter generated by the singleton containing the number!

This is often a useful approach to thinking about the more "logic-y" topological spaces: the basic open sets are usually those of the form "All points which do have property $P$", for some reasonable property $P$. If the properties we're looking at are closed under negation (e.g. in $\beta\mathbb{N}$ asking "Does $\mathcal{U}$ not contain $A$?" is the same as asking "Does $\mathcal{U}$ contain $\overline{A}$?"), then the corresponding space is totally disconnected. Sometimes this intuition just makes things messier - e.g. I don't think it's generally useful for understanding the usual topology on $\mathbb{R}$ - but other times it's quite helpful, and I tend to think that this is one of them.

Noah Schweber
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Yes, you’re correct. For each $A\subseteq\Bbb N$ let $B_A=\{\mathscr{U}\in U(\Bbb N):A\in\mathscr{U}\}$; then $\{B_A:A\subseteq\Bbb N\}$ is a base for the topology in question. Goldmakher should have put ‘for some $A\subseteq\Bbb N$’ outside the curly braces.

It’s not an easy topology to visualize, to put it mildly. As Goldmakher says, it makes $U(\Bbb N)$ the Čech-Stone compactification of $\Bbb N$ with the discrete topology, so it the universal property that characterizes the Čech-Stone compactification.

The space is zero-dimensional: $B_A=U(\Bbb N)\setminus B_{\Bbb N\setminus A}$, so it has a base of clopen sets. In fact it is even extremally disconnected, meaning the the closure of each open set is open. (Note that the word really is extremally, not extremely.) This is even stronger than being zero-dimensional (i.e., having a base of clopen sets).

Brian M. Scott
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