On the space of ultrafilters on $N$

Question

I meet the space $X$ of ultrafilters on $N$ with the topology generated by sets of the form $\{p\}\cup A$ where $A\in p \in X$. I can't understand the definition of the topology. Is the points in $N$ are all discrete? Could someone help me to understand this space? Any help will be appreciated:)

Answer 1

The points of $\Bbb N$ are isolated, so $\Bbb N$ forms a discrete open subset of $X$. $X\setminus\Bbb N$ is also discrete: if $p$ is a free (= non-principal) ultrafilter on $\Bbb N$, then $\Bbb N\in p$, so $\{p\}\cup\Bbb N$ is an open nbhd of $p$ that contains no other point of $X\setminus\Bbb N$. Thus, $X\setminus\Bbb N$ is a closed discrete subset of $X$. (And it’s a large one: there are $2^{2^\omega}=2^\mathfrak{c}$ free ultrafilters on $\Bbb N$.) However, points of $X\setminus\Bbb N$ are not isolated: $\varnothing\notin p$, so $\{p\}=\{p\}\cup\varnothing$ is not an open nbhd of $p$.

$X$ is separable, because $\Bbb N$ is a countable dense subset of $X$: every point of $X\setminus\Bbb N$ is a limit point of $\Bbb N$. However, the points of $X\setminus\Bbb N$ are limit points only of $\Bbb N$: each of them has a local base of nbhds that exclude the other points of $X\setminus\Bbb N$. Here’s a little more on the space; it’s simple stuff, but perhaps it will help you to get a better picture.

$X$ is Hausdorff.

• If $m,n\in\Bbb N$ with $m\ne n$, then $\{m\}$ and $\{n\}$ are disjoint open nbhds of $m$ and $n$.
• If $p\in X\setminus\Bbb N$ and $n\in\Bbb N$, then $\Bbb N\setminus\{n\}\in p$, so $\{n\}$ and $\{p\}\cup\big(\Bbb N\setminus\{n\}\big)$ are disjoint open nbhds of $n$ and $p$.
• If $p,q\in X\setminus\Bbb N$ with $p\ne q$, then there is some $A\subseteq\Bbb N$ such that $A\in p$ and $A\notin q$. But $q$ is an ultrafilter, so $\Bbb N\setminus A\in q$. Thus, $\{p\}\cup A$ and $\{q\}\cup\big(\Bbb N\setminus A\big)$ are disjoint open nbhds of $p$ and $q$.

Added: I should mention that the definition that you’ve been given is a little sloppy. Either $X$ should be described as the union of $\Bbb N$ and the set of free ultrafilters on $\Bbb N$, which is how I’ve treated it above, or the topology has to be defined a little differently. Specifically, if $X$ really is the set of all ultrafilters on $\Bbb N$, then the definition of the topology needs to distinguish two cases. If $p_n=\{A\subseteq\Bbb N:n\in A\}$ is the principal ultrafilter at $n\in\Bbb N$, then $\{p_n\}$ is a local base at $p_n$. Otherwise, if $p$ is free, then basic open nbhds of $p$ are the sets of the form $\{p\}\cup\{\widehat A:A\in p\}$, where $\widehat A=\{p_n:n\in A\}$.

It usually easier just to identify $p_n$ with $n$ and describe $X$ as the union of $\Bbb N$ and the set of free ultrafilters on $\Bbb N$.

Correction, 21 May 2020: My original answer said that each basic open set in $X$ is clopen, so that $X$ is zero-dimensional and hence Tikhonov. This is of course false. For instance, the closure of the basic open nbhd $\{p\}\cup\Bbb N$ of any $p\in X\setminus\Bbb N$ is clearly $X$, since $\Bbb N$ is in every free ultrafilter on $\Bbb N$. And in fact as PatrickR pointed out in this question, no basic open nbhd of a point $p\in X\setminus\Bbb N$ is closed. If $A\in p$, so that $\{p\}\cup A$ is such a nbhd, partition $A$ into two infinite subsets $B$ and $C$; exactly one of them, say $B$, belongs to $p$, and clearly every $q\in X\setminus\Bbb N$ such that $C\in q$ is in the closure of $\{p\}\cup A$. It also follows easily from this that $X$ is not regular.

Answer 2

The best interpretation I can come up with is :

$A$ represents the collection of principal ultra filters of elements in $A$. More precisely $[A] = \{[a] : a \in A\}$ where $[a] = \{S \subset N : a \in S\}$. Note that $[a]$ is an ultrafilter since for every $S$, either $a \in S$ or $a \in N - S$.

So I would say that the basic open sets are $\{p\} \cup [A] \subset X$, where $[A]$ is what I defined above.