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This is a problem I'm confused with:

Definition:

We say two subsets $A$ and $B$ of a topological space $X$ are completely separated if there exists a continuous map $f: X \rightarrow \mathbb{R}$ such that $f(A) \subseteq \{0\}$ and $f(B) \subseteq \{1\}$.

i.e we are not given that $A=f^{-1}(\{0\})$ but just that $f(A) \subseteq \{0\}$.

Prove that every pair of completely separated sets in a completely regular space have disjoint closures in their Stone–Čech compactification.

Right so we need to find a bounded continuous map $f: X \rightarrow \mathbb{R}$ and then extend this to $\beta(X)$. The problem is that how can we bound $f$ ? all we know is that there is a continuous map $f: X \rightarrow \mathbb{R}$ such that $f(x)=0$ if $x \in A$ and 1 if $x \in B$. But what if $f$ is unbounded in the points outside of $A$ and $B$. We are not given that $X = A \cup B$ so how can we bound $f$ ? can we simply define a new function say $g: X \rightarrow \mathbb{R}$ by $g(x) = min \{1,f(x)\}$ or something like that?

Thanks in advance.

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    @Andres Caicedo: Thank you very much.2011-01-20

1 Answers 1

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CW answer to push it from unanswered queue:

Let $g(x)=\min\{1,f(x)\}$, and check that $g$ is continuous. Then let $h(x)=\max\{0,g(x)\}$, and check that $h$ is continuous.

$0\le h\le1$ and then we extend it.

Then observe $A\subseteq h^{−1}(\{0\})$ and similarly $B\subseteq h^{−1}(\{1\})$.

But singletons are closed in $\mathbb R$ and $h$ is continuous so taking closures we get the result.

The proof of continuity for $h$ is the same as for $g$

(Answer is taken from comments and looks incomplete though gives an outline to proof, feel free to edit it)