Being completely normal implies perfectly normal?

Question

I know that being perfectly normal implies being completely normal. But is the converse true?

I've read somewhere that $\overline S_\omega$ is a counterexample but I don't know how to prove it.

Thanks a lot

Answer 1

The following is an outline to show that the ordinal space $\omega_1 + 1 = [ 0 , \omega_1 ]$ (this is denoted by $\overline{S}_\Omega$ in Munkres's Topology) of all countable ordinals together with the least uncountable ordinal under the order topology is completely normal but not perfectly normal. An important thing about this space is that for each $\beta \in [0,\omega_1]$ the family of all sets of the form $$( \alpha , \beta ] = \{ \xi \in [0,\omega_1] : \alpha < \xi \leq \beta \}$$ for $\alpha < \beta$ is a neighbourhood base at $\beta$.

• It is completely normal because all linearly ordered spaces are completely normal. Okay, that might be too short of an outline. For this particular space, suppose that $A, B \subseteq [0,\omega_1]$ are nonempty separated subsets (i.e., $A \cap \overline{B} = \varnothing = \overline{A} \cap B$). Without too much loss of generality we may assume that $0$ does not belong to either set (it is an isolated point of the space).

  • For each $\alpha \in A$ find a $\gamma_\alpha < \alpha$ such that $( \gamma_\alpha , \alpha ] \cap B = \varnothing$, and then set $U = \bigcup_{\alpha \in A} ( \gamma_\alpha , \alpha ]$.
  • Similarly, for each $\beta \in B$ find a $\delta_\beta < \beta$ such that $( \delta_\beta , \beta ] \cap A = \varnothing$. Then set $V = \bigcup_{\beta \in B} ( \delta_\beta , \beta ]$.

  Show that these are disjoint open neighbourhoods of $A$, $B$, respectively.

• It is not perfectly normal because $\{ \omega_1 \}$ is a closed subset which is not G$_\delta$. (To show that if $\{ U_n \}_{n \in \mathbb{N}}$ is a family of open neighbourhoods of $\omega_1$, then $\bigcap_{n \in \mathbb{N}} U_n \neq \{ \omega_1 \}$ note that if $\{ \alpha_n \}_{n \in \mathbb{N}}$ is a countable family of countable ordinals, then $\sup_{n \in \mathbb{N}} \alpha_n$ is also a countable ordinal. For each $n$ pick $\alpha_n < \omega_1$ such that $( \alpha_n , \omega_1 ] \subseteq U_n$.)

Answer 2

Being that definitions relating to separation tend not to agree, I thought it prudent to add the definitions used in $\pi$-Base. They are the ones found in Willard's General Topology, which seems to be fairly standard.

A completely normal space is one for which any subspace is normal (under the subspace topology). Equivalently, a space is completely normal provided any two separated sets can be placed into disjoint open sets.

A space $X$ is perfectly normal if, for distinct closed sets $A$ and $B$, there is a continuous function $f: I \rightarrow \mathbb{R}$ such that $f^{-1}(A) = \{0\}$ and $f^{-1}(B) = \{1\}$.

$\pi$-Base, an online version of Steen and Seebach's Counterexamples in Topology, lists the following spaces as completely normal but not perfectly normal. You can view the search result to learn more about these spaces.

Alexandroff Square

Baire Product Metric on Rω

Closed Ordinal Space [0,Ω]

Concentric Circles

Countable Excluded Point Topology

Deleted Integer Topology

Either-Or Topology

Finite Excluded Point Topology

Fortissimo Space

Hjalmar Ekdal Topology

Indiscrete Topology

Lexicographic Ordering on the Unit Square

Nested Interval Topology

Odd-Even Topology

One Point Compactification Topology

Open Ordinal Space [0,Ω)

Right Order Topology on R

Sierpinski Space

The Extended Long Line

The Integer Broom

The Long Line

Uncountable Excluded Point Topology

Uncountable Fort Space