Stone-Čech Compactification of The integers is $T_5$?

Question

Since $\beta\mathbb{N}$ is $T_2$ and compact we can conclude that it is $T_4$. However, what about $T_5$? My only guess will be that $\beta\mathbb{N}-\mathbb{N}$ is not normal. Is this true?

Answer 1

No, as all points of $\mathbb{N}$ are isolated ,it's an open subset of $\beta\mathbb{N}$ (in general any locally compact Hausdorff space $X$ will be open in $\beta X$). So its remainder $\beta\mathbb{N} \setminus \mathbb{N}$ is a closed subset of $\beta \mathbb{N}$ hence compact + $T_2$ thus normal.

this paper has a construction of non-normal subspaces.

Answer 2

Streamlined version of the example in the reference that I gave in a comment:

(1). Let $f:\mathbb Q\to \mathbb N$ be a bijection. For $r\in \mathbb R$ let $(s_r(j))_{j\in \mathbb N}$ be a sequence of rationals in $(r,\infty)$ that converges to $r.$ Let $T(r)=\{f(s_r(j)):j\in \mathbb N\}.$

This is to obtain a family $\{T(r):r\in \mathbb R\}$ of infinite subsets of $\mathbb N$ such that $T(r_1)\cap T(r_2)$ is finite whenever $r_1\ne r_2.$

(2). We identify $\beta \mathbb N$ with the Wallman extension $w\mathbb N,$ which is the space of ultra-filters on $\mathbb N.$ For $n\in \mathbb N$ we identify the fixed ultra-filter $\{S\subset \mathbb N: n\in S\}$ with $n$ itself. A basic open set in $\beta \mathbb N$ is $$U^*=\{f\in w\mathbb N: U\in f\}$$ for any $U\subset \mathbb N.$ Observe that $$ U_1^*\cap U_2^* =(U_1\cap U_2)^*$$ so if $U_1\cap U_2$ is finite then $(U_1^*\cap U_2^*)\cap (w\mathbb N$ \ $\mathbb N)=\phi.$

(3).For each $T(r)$ in part (1) above choose $G(r)\in (T(r))^*$ \ $\mathbb N.$ Let $S=\{G(r): r\in \mathbb N\}\cup \mathbb N.$

(i). $\mathbb N$ is dense in $w\mathbb N,$ and $\mathbb N\subset S$, so $\mathbb N$ is dense in the subspace $S.$ So $S$ is separable.

(ii). $\mathbb N$ is discrete, hence locally compact, hence $\mathbb N$ is an open subset of $w\mathbb N.$ So $\mathbb N$ is an open subset of the subspace $S.$

(iii). $( T(r))^*$ is a nbhd of $G(r)$ in $w\mathbb N$ that does not contain $G(r')$ for any $r'\ne r.$ So in $S$ the set $A=\{G(r):r\in \mathbb R\}$ is a discrete subspace of $S$. And by (ii), $A$ is closed in $S.$

(4). A separable space $S$ with a closed discrete subspace $A$, where the cardinal of $A$ is the cardinal of $R,$ cannot be a normal space.