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I was browsing Gillman and Jerison for known relations between zero-sets, $C$-embedded sets and so on.

The spaces I'm considering are $T_{3.5}$.

There are two properties that pseudocompact spaces have

  1. All countable $C$-embedded sets are compact

  2. All countable zero-sets are compact

Where (1) is equivalent to pseudocompactness, but (2) isn't.

A countable closed discrete $C^*$-embedded $G_\delta$-set in a pseudocompact space doesn't have to be either a zero-set nor $C$-embedded, as Katetov's example $\mathbb{N}\subseteq\Lambda = \beta\mathbb{R}\setminus (\beta\mathbb{N}\setminus\mathbb{N})$ shows.

A zero-set need not be $C^*$-embedded, nor does a $C$-embedded set need to be closed.

A zero-set is $C^*$-embedded iff $C$-embedded.

However, a countable $C$-embedded set is closed.

Must a countable zero-set be $C$-embedded?

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2 Answers 2

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Most of the following can be found in Gillman and Jerison.

Lemma Subsets $A,B\subseteq X$ are completely separated if and only if they are contained in disjoint zero-sets. $\quad\blacksquare$

A subset $A\subseteq X$ is well-embedded if it is completely separated from any disjoint zero-set.

Corollary Every zero-set is well-embedded. $\quad\blacksquare$

A subspace $A\subseteq X$ is z-embedded if every zero-set of $A$ is the trace of a zero-set of $X$. Every C$^*$-embedded subspace is z-embedded. The following result is found in the paper On the Structure of a Class of Archimedian Lattice-Ordered Algebras by Henriksen and Johnson, where it is attributed to Jerison.

Proposition A Lindelöf subspace of a Tychonoff space is z-embedded. $\quad\blacksquare$

In fact, there is a kind of converse for this last statement: a Tychonoff space $X$ is z-embedded in every Tychonoff space containing it if and only if $X$ is either Lindelöf or almost-compact.

Finally we have the following, which I believe is due to Blair and Hager.

Proposition A subspace $A\subseteq X$ is C-embedded if and only if it is both z-embedded and well-embedded.

This gives us the following.

Corollary Any z-embedded zero set is C-embedded. $\quad\blacksquare$

Corollary If $X$ is a Tychonoff space and $A\subseteq X$ is a Lindelöf zero-set, then $A$ is C-embedded in $X$. $\quad\blacksquare$

Corollary If $X$ is a Tychonoff space, then every countable zero-set of $X$ is C-embedded. $\quad\blacksquare$

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    $\begingroup$ I think the paper Extensions of Zero-Sets and of Real-Valued Functions by Blair and Hager collects a few statements and has a decent bibliography. A lot of papers cite the text book Normal Topological Spaces by Alò and Shapiro as general reference for z-embeddings, but the book seems to be pretty rare and I've never seen a copy myself. $\endgroup$
    – Tyrone
    Commented Jul 1 at 11:26
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    $\begingroup$ I never realised before that Blair didn't use the term, as it's fairly standard now. I guess I must have picked it by reading K. Yamazaki's papers. Apparently the terminology was introduced in Moran's 1970 paper Measures on Metacompact Spaces. $\endgroup$
    – Tyrone
    Commented Jul 1 at 13:03
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    $\begingroup$ Perhaps its interesting to point out that corollary 4.2 from Extensions of Zero-Sets and of Real-Valued Functions by Blair and Hager also gives us a partial converse i.e. a Lindelöf C-embedded set is closed. This is a particular case of the fact that Lindelöf spaces are realcompact and a realcompact $C$-embedded set is closed. $\endgroup$
    – Jakobian
    Commented Jul 11 at 13:31
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    $\begingroup$ Or another: a Lindelöf subspace of a Tychonoff space is a zero-set if and only if it is a C-embedded $G_\delta$-subset. $\endgroup$
    – Tyrone
    Commented Jul 11 at 15:04
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    $\begingroup$ Call $S$ strongly well-embedded if $S$ is completely separated from each disjoint closed set. This property implies $S$ is closed, and a zero-set if its $G_\delta$. A space $X$ is normal iff every closed subspace is strongly well-embedded. And a Lindelöf well-embedded subspace is strongly well-embedded. One of the implications in above comment follows from these facts. $\endgroup$
    – Jakobian
    Commented Jul 13 at 12:45
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All countable zero-sets are $C$-embedded.

I'll need the following from the book "Rings of continuous functions" by Gillman and Jerison.

Completely separated: Sets $D_1, D_2$ are completely separated if there exists disjoint zero sets $D_1\subseteq Z_1, D_2\subseteq Z_2$.

Theorem 1.17: If $Y\subseteq X$, then $Y$ is $C^*$-embedded in $X$ iff completely separated sets in $Y$ are completely separated in $X$.

Exercise 3B.1: If $D$ is countable and $F$ is closed, $D\cap F = \emptyset$, then there exists a zero-set $Z$ with $D\cap Z = \emptyset$ and $Z\supseteq F$.

Exercise 1F.1: A $C^*$-embedded zero-set is $C$-embedded


Let $D$ be a countable zero-set and $D_1, D_2\subseteq D$ be disjoint zero-sets of $D$. Thus $D_1, D_2$ are disjoint countable closed sets in $X$. From exercise 3B.1 there exists a zero set $Z_1\supseteq D_1$ of $X$ with $Z_1\cap D_2 = \emptyset$, and we can assume $Z_1\subseteq Z$ by replacing $Z_1$ with $Z_1\cap Z$ since $Z$ is a zero-set. Similarly, since $Z_1$ is countable, we can now find a zero-set $Z_2$ with $Z_1\cap Z_2 = \emptyset$ and $Z_2\supseteq D_2$. Thus $D_1, D_2$ are completely separated in $X$. From Urysohn's lemma 1.17, $D$ is $C^*$-embedded, hence $C$-embedded.

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