Is such space separable?

Question

Here is a kind of space which called Katětov extension of the natural number from this remark:

I have two questions on this space:

1 Is such space separable?

2 Is such space first countable?

Thanks very much.

Answer 1

Seirios has already given an excellent full answer. Here’s a more hands-on answer to the second question. The Katětov extension of $\Bbb N$ is

$$\kappa\Bbb N=\Bbb N\cup\{p\subseteq\wp(\Bbb N):p\text{ is a free ultrafilter on }\Bbb N\}\;,$$

where a local base at $p\in\kappa\Bbb N\setminus\Bbb N$ is $\mathscr{B}_p=\big\{\{p\}\cup U:U\in p\big\}$; it is not first countable at any point of $\kappa\Bbb N\setminus\Bbb N$.

Fix $p\in\kappa\Bbb N\setminus\Bbb N$. Suppose that $\{U_n:n\in\Bbb N\}\subseteq\mathscr{B}_p$. Recursively choose $m_k,n_k\in\Bbb N$ for $k\in\Bbb N$ so that $m_k\ne n_k$ and

$$m_k,n_k\in U_k\setminus\big(\{m_i:i<k\}\cup\{n_i:i<k\}\big)$$

for each $k\in\Bbb N$. Let $M=\{m_k:k\in\Bbb N\}$, and let $N=\Bbb N\setminus M$. Then for each $k\in\Bbb N$ we have $n_k\in U_k\setminus M$ and $m_k\in U_k\setminus N$, so $U_k\nsubseteq M$ and $U_k\nsubseteq N$ for all $k\in\Bbb N$. But $p$ is an ultrafilter, so either $M\in p$ or $N\in p$; without loss of generality $M\in p$, and $\{p\}\cup M$ is then an open nbhd of $p$ that does not contain any of the sets $U_k$ for $k\in\Bbb N$. Thus, $\{U_k:k\in\Bbb N\}$ is not a base at $p$, and $\kappa\Bbb N$ is not first countable at $p$.

Answer 2

Of course, $\mathbb{N}$ is dense in $\kappa \mathbb{N}$, so $\kappa \mathbb{N}$ is separable. Moreover, any separable first countable space has cardinality at most $2^{\omega}$ (any point is the limit of a sequence of the countable dense subspace), whereas $|\kappa \mathbb{N}|=2^{2^{\omega}}$.