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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.

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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 $X$.

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

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