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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-Cech 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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6  
Sure. Let $g(x)={\rm min}\{1, f(x)\}$, and check that $g$ is continuous. Then let $h(x)={\rm max}\{0,g(x)\}$, and check that $h$ is continuous. –  Andres Caicedo Jan 20 '11 at 7:07
    
You could proceed by contradiction. –  PEV Jan 20 '11 at 7:18
1  
@Andres Caicedo: I think I get it, so $0 \leq h \leq 1$ 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. Is this alright? thank you again. –  student Jan 20 '11 at 7:20
    
Exactly. Of course, the proof of continuity for $h$ is the same as for $g$. –  Andres Caicedo Jan 20 '11 at 7:23
    
@Andres Caicedo: Thank you very much. –  student Jan 20 '11 at 7:27

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