Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 / A / 2

Urysohn's lemma states that if $X$ is normal and $A,B$ are disjoint closed subsets, there is a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. The values on the rest of $X$ remain in that interval. Interchanging $A,B$ or composing with an affine function gives any two prescribed real endpoint values. This is functional separation of closed sets, stronger than merely choosing disjoint open neighbourhoods.

A C-embedded subset $Y\subseteq X$ is one when every continuous real-valued function on $Y$ extends continuously to $X$. It is a Cb-embedded subset when this is required for every bounded continuous real-valued function, with a bounded extension. These definitions concern the subspace topology on $Y$.

Here is the bounded Tietze extension construction from Urysohn's lemma. Let $g:Y\to \mathbb{R}$ be continuous with $|g|\le M$. The case $M=0$ has the zero extension. For $M>0$, the sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are disjoint and closed in $X$, since $Y$ is closed. Urysohn's lemma gives a continuous $u_0:X\to [-M/3,M/3]$ taking the endpoint values on $A,B$. On $Y$,Indeed on $A$ or $B$ this follows from the prescribed endpoint; on their complement both $|g|$ and $|u_0|$ are at most $M/3$.

Repeat on the residual. With $M_n=M(2/3{)}^n$, choose continuous $u_n$ satisfying $\Vert u_n{\Vert }_{\infty }\le M_n/3$ and reducing the residual on $Y$ to at most $M_{n+1}$. The series $F={\sum }_{n\ge 0}{u}_n$ converges uniformly by the geometric majorant. Its sum is continuous, has $|F|\le \sum M_n/3=M$, and restricts to $g$, since the residual tends uniformly to zero. Thus every bounded continuous function extends, with its bound preserved, proving $C^b$-embeddedness without assuming the extension theorem being proved.

For a strict-range modification of a continuous extension, the closed sets $Y$ and $Z=\{x:|f(x)|\ge 1\}$ are disjoint. Choose by Urysohn's lemma a continuous $u:X\to [0,1]$ with $u{|}_Y=0$ and $u{|}_Z=1$, and defineThis function is continuous and agrees with $f$ on $Y$. On $Z$ it vanishes. Off $Z$ we have $|f|<1$ and $0\le 1-u\le 1$, so $|\tilde{f}|\le |f|<1$. Hence its range lies strictly inside $(-1,1)$ on all of $X$. The construction also covers empty $Y$ or empty $Z$ by a constant separating function where appropriate.

Let $h:Y\to \mathbb{R}$ be continuous, without a boundedness assumption. The homeomorphism $\theta (t)=t/(1+|t|)$ maps the real line onto $(-1,1)$, with inverse ${\theta }^{-1}(s)=s/(1-|s|)$. The bounded function $g=\theta \circ h$ extends by the preceding bounded construction to a continuous real-valued $G$ on $X$. Since $|G|<1$ on $Y$, the strict-range modification gives a continuous $\tilde{G}$ with range in $(-1,1)$ and the same restriction to $Y$. Thereforeis continuous everywhere: its denominator is pointwise positive. On $Y$ it equals ${\theta }^{-1}(\theta (h))=h$. Thus $Y$ is $C$-embedded. It is important to modify the extension before applying the inverse: an extension allowed to attain $1$ or $-1$ outside $Y$ would make this formula undefined there.

Take the natural numbers as the dense discrete subspace of their Stone-Čech compactification $\beta \mathbb{N}$. The ambient compact Hausdorff space is normal, and the Stone-Čech extension property makes this subspace $C^b$-embedded. It is not $C$-embedded: the continuous function $n\mapsto n$ is unbounded, whereas every continuous real-valued function on a compact space is bounded. The subspace is not closed, so this does not contradict the preceding closed-subspace theorem.