# How can I show that $\beta\mkern1.5mu \Bbb N$ is not a hereditarily normal space?

Show that $$\beta\mkern1.5mu \Bbb N$$ (the Stone Cech compactification of natural numbers) isn't hereditarily normal. I was thinking about defining a continuous and closed function on some non-normal space into $$\beta \Bbb N$$. But I don't know which one could work.

Recall the following fact about the Stone-Čech compactification of normal spaces:

Fact. If $$X$$ is a normal space, and $$E, F \subseteq X$$ are disjoint closed subsets of $$X$$, then $$\overline{E} \cap \overline{F} = \emptyset$$ where the closure is taken in $$\beta X$$.

Coro. If $$X$$ is a normal space, and $$U \subseteq X$$ is clopen, then $$\overline{U}$$ is clopen in $$\beta X$$.

A very slight extension of the Fact for the discrete space $$\mathbb{N}$$ yields the following.

Coro. If $$A , B \subseteq \mathbb{N}$$ are such that $$A \cap B$$ is finite, then $$\overline{A} \cap \overline{B} \subseteq A \cap B \subseteq \mathbb{N}$$ where the closure is taken in $$\beta \mathbb{N}$$.

Fact. If $$A \subseteq \mathbb{N}$$ is infinite, then $$\overline{A} \not\subseteq \mathbb{N}$$ where the closure is taken in $$\beta \mathbb{N}$$.

To get on with showing that $$\beta \mathbb{N}$$ is not hereditarily normal, fix a family $$\{ A_i : i \in \mathbb{R} \}$$ is infinite subsets of $$\mathbb{N}$$ such that $$A_i \cap A_j$$ is finite for distinct $$i,j \in \mathbb{R}$$.

For each $$i \in \mathbb{R}$$ fix some $$x_i \in \overline{A_i} \setminus \mathbb{N}$$, and define $$D := \{ x_i : i \in \mathbb{R} \} \subseteq \beta \mathbb{N} \setminus \mathbb{N}$$. Using the above we can show that $$D$$ is a discrete subspace of $$\beta \mathbb{N}$$ of cardinality $$| \mathbb{R} | = 2^{\aleph_0}$$.

Now set $$Y := D \cup \mathbb{N}$$. We can show that $$\mathbb{N}$$ is a dense subset of $$Y$$, and that $$D$$ is a closed subset of $$Y$$.

By Jones's Lemma (together with Cantor's Theorem $$2^\kappa > \kappa$$), it follows that $$Y$$ cannot be normal:

Jones's Lemma. If $$X$$ is a separable normal space, then $$2^{|D|} \leq 2^{\aleph_0}$$ for every closed discrete subspace $$D$$ of $$X$$.

• Thank you ver much I was realy stuck in that one. – Ramon Poo Ramos Nov 3 '19 at 1:01