$G_{\delta }$ Sets in $\beta (\mathbb N)\setminus \mathbb N.$

Question

I am working on a past prelim exercise, and my proof seems to be too long and dfficult for such an exercise. I would appreciate feedback.

Let $t\in \beta (\mathbb N)\setminus \mathbb N.$ Then, $\left \{ t \right \}$ is not a $G_{\delta}$. The hint is to consider the clopen sets in $\beta \mathbb N$ and use the fact that $|\beta (\mathbb N)|=2^c.$

Here is my attempt:

$1).$ As a first step, I have been able to show that the clopen sets in $\beta (\mathbb N)$ are exactly the closures of the open sets in $\mathbb N, $ and that these form a basis for $\tau_{\beta (\mathbb N)}.$

$2).$ So that, if $\left \{ t \right \}$ is a $G_{\delta}, $ then there is a decreasing sequence of sets $\varnothing \neq E_n\subseteq \mathbb N$ such that $E_n$ are open in $\mathbb N$, and $\overline E_n$ are clopen (in $\beta (\mathbb N)$) and $\bigcap_n \overline E_n=\left \{ t \right \}.$

Since the $E_n$ are decreasing and non-empty, we may choose distinct points $x_n\in E_n$ and so we obtain the sequence $(x_n)_{n\in \mathbb N}.$

$3).$ I claim that $\overline { (x_n) }\setminus \mathbb N\subseteq \bigcap_n \overline E_n:$

Let $x\in \overline { (x_n) }\setminus \mathbb N$ and suppose not. Then, there is an $N\in \mathbb N$ such that $x\notin \overline E_N.$ This implies that there is an open $U$ containing $x$ such that $U\cap E_N=\varnothing.$ As the $E_n$ are decreasing this means that $\left \{ x_N,x_{N+1},\cdots \right \}\cap U=\varnothing.$ On the other hand if $x\notin \overline E_N, $ then of course $x\in \overline E_N^c, $ which is open. Now we may conclude that $U\cap \overline E_N^c$ is an open set containing $x$ which does not intersect $( x_n), $ a contradiction.

$4).$ To finish, note that $(x_n)\subseteq \mathbb N$ is infinite, and so homeomorphic to $\mathbb N$ which means that $\overline { (x_n) } $ is homeomorphic to $\beta (\mathbb N), $which is impossible since, by $3).$, $|\overline { (x_n) } |=\omega.$

Answer 1

Your proof is mostly fine (but see below), and as far as I know is essentially the simplest way to prove this statement.

The hint to use that $|\beta(\mathbb{N})|=2^c$ seems very strange to me; I don't see any natural way to use it. You have used it only in a very weak way in step 4 (where all you actually needed was the fact that $|\beta(\mathbb{N})\setminus\mathbb{N}|>1$, so that $\overline{(x_n)}\setminus\mathbb{N}$ must contain some point other than $t$).

Here's what I would guess whoever wrote the problem had in mind. If you only wish to show that $\{t\}$ is not a $G_\delta$ for some $t$, rather than for every $t$, there is a simple argument that does use the fact that $|\beta(\mathbb{N})|=2^c$. Namely, note that since there are only $c$ subsets of $\mathbb{N}$, there are only $c^{\aleph_0}=c$ different sequences of subsets $(E_n)$ such as the one you consider in step 2. This means that there can only be at most $c$ different $G_\delta$ singletons in $\beta(\mathbb{N})$, since each one must be the intersection of some such sequence $(E_n)$. Since $2^c>c$, some singleton must not be $G_\delta$.

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OK, now let me nitpick your proof. In step 2 you should mention that the $E_n$ are infinite to be sure you can really find a sequence of distinct points. (If $E_n$ were finite, then $\overline{E_n}=E_n$ so $t\in E_n\subset\mathbb{N}$ giving a contradiction.)

In step 3, to be sure your open neighborhood of $x$ really is disjoint from $(x_n)$, you also need to remove the finitely many points $x_1,x_2,\dots,x_{N-1}$ from it. (Here you use the assumption that $x\not\in\mathbb{N}$, so $x$ cannot be equal to any of those points.) Note that it is also unnecessary to intersect $U$ with $\overline{E_N}^c$: since $U$ is an open set that is disjoint from $E_N$, it must automatically also be disjoint from the closure.

In step 4, it is not obvious that just because $(x_n)$ is homeomorphic to $\mathbb{N}$, its closure is homeomorphic to $\beta(\mathbb{N})$. All you know a priori is that its closure is some compactification of $\mathbb{N}$. To prove that it is homeomorphic to $\beta(\mathbb{N})$, you could use the fact that every continuous function $(x_n)\to[0,1]$ extends continuously to $\mathbb{N}$ and hence to $\beta(\mathbb{N})$, and hence in particular it extends to $\overline{(x_n)}$.