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Let $ X$ be a topological space. The set $C(X)$ is all continuous, real valued functions on $X$, so that $C(X) \subseteq \mathbb{R}^{X}$.

Define:

$C^* (X) = \{ f \in C(X) \vert \quad f\quad is \quad bounded \}$.

When $C^* (X) = C(X)$, $X$ is said to be pseudocompact. Every compact space is pseudocompact.

1:A discrete zero-set is $C^{*}$-embedded $\Longleftrightarrow$ all of its subsets are zero-set.

I proved the necessary part, To show all of its subsets are zero-set, should we use Urysohn’s extension theorem?How?

Urysohn’s extension theorem: A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are completely separated in $X$.

Another my question is:

2: If, for any two disjoint zero-sets in $X$, at least one is compact , or even pseudocompact, then $X$ is pseudocompact.

[ If $C = C^{*}$, then some function $f$ in $C$ assume the values $0$ and $1$ infinitely often on a $C$-embedded copy of $\mathbb{N}$]

Henno Brandsma
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Jak
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1 Answers1

2

First 1A. It follows from basic knowledge on zero-sets, essentially:

Let $S$ be a discrete zero-set. Suppose it is $C^\ast$-embedded. Let $A \subseteq S$ be any subset, then define the function $f_A: S \to [0,1]$ by $f(x) = 0 $ iff $x \in A$ and $f(x) = 1$ for $x \in S\setminus A$. Then $f_A$ is continuous as $S$ is discrete (so any function on it is continuous) and as $S$ is $C^\ast$-embedded in $X$ we have an extension $\hat{f}_A: X \to [0,1]$. But then $A= S \cap \hat{f}_A^{-1}[\{0\}]$ is a sero-set as a finite intersection of zero sets of $X$. So all subsets of $S$ are zero-sets.

If all subsets of $S$ are zero-sets, then in particular $S$ itself is a zero-set and all its subsets are closed (zero-sets are closed) so $S$ is discrete as well. To see it is $C^\ast$-embedded, take two completely separated subsets $A, B \subseteq S$, and by assumption these are zero-sets so $A = f^{-1}[\{0\}]$, and $B = g^{-1}[\{0\}]$ for continuous $f,g : X \to [0,1]$ but then $A$ and $B$ are completely separated in $X$ (in fact disjoint zero-sets always are) by $h = \frac{f}{f+g}$ e.g. So Tietze-Urysohn implies that $S$ is $C^\ast$-embedded.

As to 2. assume that for any two disjoint zero-sets at least one of them is pseudocompact. Then we must show $C$ is pseudocompact.

Assume not, then there is an unbounded continuous $f: X \to \mathbb{R}$. Then we have points $x_n \in X$ with $|f(x_n)| > n$. The argument I gave here shows that $A = \{x_n: n \in \mathbb{N}\}$ is a closed and discrete subspace of $X$. Your hint seems to suggests that $A$ is a $C$-embedded copy of $\mathbb{N}$, and this is indeed the case by Gilman and Jerison, 1.19 which proves that $A$ is a $C$-embedded copy of $\mathbb{N}$. (sketch of proof: $h$, defined as $f$ restricted to $A$, is a homeomorphism between $A$ and a closed discrete subset $f[A]$ of $\mathbb{R}$, with inverse $h'$. So if $g$ is any real-valued function on $A$, we can extend $g \circ h'$ from $f[A]$ to $\mathbb{R}$ and then $f$ composed with this extension extends $g$ to $X$.)
So we can define a $g$ on $A$ that is $0$ on even-indexed points and $1$ for the odd-indexed ones, and extend it to $g'$ on $X$ and then we have two disjoint zero-sets (the sets $g'^{-1}[\{0\}]$ and $g'^{-1}[\{1\}]$ which both are not pseudocompact (as witnessed by $f$), contradicting the assumption on $X$. Hence $X$ is pseudocompact.

Henno Brandsma
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