Necessary and sufficient condition for a set in $\mathbb{R}$ to be C-embedded or $C^*$-embedded

Question

I saw the following result as an exercise in the book by Gillman and Jerison Rings of continuous functions.The statement says

Consider the topological space $(\mathbb{R}, \tau_{u})$ where $\tau_{u}$ is the usual topology of $\mathbb{R}$ and $S\subset \mathbb{R}$. Then $S$ is $C$-embedded in $\mathbb{R}$ (resp. $C^*$-embedded) if and only if $S$ is closed in $\mathbb{R}$

I already show that if $S$ is a closed set of $\mathbb{R}$ then $S$ is $C$-embedded in $\mathbb{R}$. Since $(\mathbb{R}, \tau_{u})$ is a $T_{4}$ space. I use the following result

Let $(X, \tau)$ be a topological space, then the following conditions are equivalent:

1. $X$ is a $T_{4}$ space.
2. Two closed sets of $X$ disjoint, are completely separated in $X$.
3. Every closed set of $X$ is $C$-embedded in X.
4. Every closed set of $X$ is $C^*$-embedded in X.

The exercise gives a suggestion to solve the part of supposing that if $S$ is $C$-embedded in $\mathbb{R}$ then it is closed in $\mathbb{R}$. The suggestion is to suppose that $S$ is not closed in $\mathbb{R}$ and to see that the set $\overline{S}-S$ is non-empty, which implies that there exists at least one point in $\overline{S}-S$, then there exists a sequence in $S$ that converges at that point, thus arriving at a contradiction that $S$ is $C$-embedded in $\mathbb{R}$.

I am trying to show that if $S$ is $C$-embedded in $\mathbb{R}$ then it is closed in $\mathbb{R}$, but without using the suggested sequence in the problem, I know the following results for $T_{4}$ spaces:

• $(X, \tau)$ is a space $T_{4}$ if and only if each closed subset of $X$ is $Z$-embedded in $X$.

• Let $(X, \tau)$ be a topological space, then the following conditions are equivalent.

  1. $X$ is a $T_{4}$ space.
  2. Each generalized-$F_{\sigma}$ set in $X$ is $Z$-embedded in $X$.
  3. Each $F_{\sigma}$ set on $X$ is $Z$-embedded on $X$.
  4. If $S$ is a closed subset of $X$ and $Z$ is a zero-set of $X$, then $S\cup Z$ is $Z$-embedded in $X$.
  5. If $S$ is a closed subset of $X$ and $Z$ is a zero-set of $X$, such that $S\cap Z=\emptyset$ then $S\cup Z$ is $Z$-embedded in $X$.
  6. Each closed subset of $X$ is $Z$-embedded in $X$.

But still I can not prove that if $S$ is $C$-embedded in $\mathbb{R}$ then it is closed in $\mathbb{R}$, if anyone has any suggestions on how to do it please give me a clue.

Answer 1

Each $C$-embedded subset $S$ of a metrizable space $X$ is closed, otherwise we would pick a point $x_0\in \overline{S}\setminus S$ and a metric $d$ on the space $X$ and put $f(x)=1/d(x,x_0)$ for each $x\in S$. Since the function $f$ is unbounded at each neighborhood ot the point $x_0$, it cannot be extended to a continuous function from $S\cup\{x_0\}$ to $\Bbb R$, a contradiction.

Althoug the above claim answers your question, below I provide a more strong claim for those who may need it.

Proposition. Each $C^*$-embedded subset $S$ of a first countable Tychonoff space $X$ is closed.

Proof. Let $S$ be a non-closed subset of the space $X$. Pick a point $x_0\in \overline{S}\setminus S$ and a countable base $\{U_n\}$ of open neighborhoods of the point $x_0$. For each $n$ take a continuos function $f_n:X\to [0,1]$ such that $f(x_0)=1$ and $f_n(X\setminus U_n)=\{0\}$, and put $f(x)=\sum f_n(x)$ for each $x\in S$. It is easy to check that the function $f$ is correctly defined and continuous on $S$ and the image $f(S)\subset\Bbb R$ is unbounded. So it contains a sequence $Y=\{y_n\}$ of distinct points such that $|y_n|>n$ for each $n$. Define a function $g:Y\to [0,1]$ such that $g(y_n)=0$ for even $n$ and $g(y_n)=0$ for odd $n$. Since $Y$ is a closed discrete subspace of a normal space $\Bbb R$, the function $g$ can be extended to a continuous function $\hat g:\Bbb R\to [0,1]$. The construction implies that the function $\hat gf:S\to [0,1]$ cannot be extended to a continuous function from $S\cup\{x_0\}$ to $[0,1]$, so the set $S$ is not $C^*$-embedded in the space $X$. $\square$