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I want to show the set of ultrafilters $\beta \mathbb{N} - \mathbb{N}$ (where $\beta \mathbb{N}$ is all ultrafilters on $ \mathbb{N}$) is not separable. I know we can take as a base of $\beta \mathbb{N}$ the sets $c_A = \{U \in \beta \mathbb{N}: A \in U\}$; so I think it is necessary and sufficient for a dense set of ultrafilters $(U_n)$ to be dense in the $c_A$: i.e. every $c_A$ contains some $U_i$.

So, I think to prove nonseparability I should start by assuming there is such a set of $U_n$ with every $c_A$ containing some element of the set, then somehow deduce a contradiction. I would prefer not to use too many preliminary results if a reasonably direct proof is possible, though I know that for any ultrafilter $U$, for any set $A$ either $A$ or $A^c \in U$, and I know $\beta \mathbb{N}$ is compact Hausdorff with the topology defined by the above base. Could anyone help me please? I tried cleverly selecting the sets A in the base of $c_A$ but I haven't had any luck deriving a contradiction.

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Every $c_A$ (for $A$ infinite) contains some $U_n$ iff every (infinite) $A$ belongs to some $U_n$. Derive a contradiction by finding some $A$ which belongs to no $U_n$. Do this by diagonalizing against the $U_n$. Naively, you might want to pick an $A_n$ not in $U_n$, then intersect the $A_n$ to get $A$, but $A$ might be finite. Indeed if we just choose haphazardly, $A_0$ and $A_1$ might be disjoint. A slightly more prudent approach would be to build $A$ recursively, choosing $A_0 \notin U_0$, then splitting $A_0$ into two infinite sets, and choosing $A_1$ to be any one of the two which isn't in $U_1$ (this will have to be true of at least one of the two pieces of $A_0$ since $U_1$ is an ultrafilter), and so on. We could then take the intersection of the $A_n$, but this intersection may still end up finite.

What we'll do is something like the above, but at each stage we'll pick some number $a_n$ and lock it into $A$, so that we're sure $A$ is infinite in the end. The resulting $A$ might not be a subset of all of our $A_n$, so how will we know that $A \notin U_n$ for each $n$? The point will be that even if $A \not\subset A_n$, we will have $A \subset^\ast A_n$, that is, $A$ will be a subset of each $A_n$ "mod finite," and since the ultrafilters $U_n$ are non-principal, this will be good enough.

We will build the $A_n$ recursively. Fix an infinite $A_0 \notin U_0$. Let $a_0 = \min A_0$. Fix $A_{n+1} \subset A_n$ infinite such that $\min A_{n+1} > a_n$ and $A_{n+1} \notin U_{n+1}$. Let $a_{n+1} = \min A_{n+1}$. Set $A = \{a_n : n < \omega\}$.

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Hi Amit, many thanks for your extremely helpful answer. That makes perfect sense to me except for the final step in the logic: arguing that $A \not \in U_n$ because $A \subset ^* A_n$ and the $U_n$ are non-principal. Is there perhaps some result I don't know of where if one set A differs from another, B, by only finitely many elements with a non-principal ultrafilter U and B is not contained in U then A is not contained in U? My understanding of a principal ultrafilter $\tilde{x}$ is that it is the set of all subsets of $\mathbb{N}$ which contain a specified element $x$. –  Thomas Nov 21 '11 at 10:53
    
Is this because every non-principal ultrafilter extends the cofinite filter and so we could take an intersection with the complement of 'the things in $A$ not in $A_n$' to obtain a subset of $A_n$ in $U_n$, and deduce that $A_n \in U_n$ for contradiction? –  Thomas Nov 21 '11 at 11:21
    
@Thomas: Yes. If $\mathscr{U}$ is a free ultrafilter on $\mathbb{N}$, and $A,B\subseteq\mathbb{N}$ differ by a finite set, then $A\in\mathscr{U}$ iff $B\in\mathscr{U}$, basically for the reason that you suggest: choose a cofinite $C\in\mathscr{U}$ s.t. $C\cap A=C\cap B=A\cap B$, and note that $A\in\mathscr{U}$ iff $A\cap C\in\mathscr{U}$ iff $B\in\mathscr{U}$. –  Brian M. Scott Nov 21 '11 at 12:23
    
Brilliant, thank you for the help! –  Thomas Nov 21 '11 at 14:00

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