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A completely regular space X is an F-space if for each functionally open set $M\subset X$ every continous function $f: M \rightarrow I $ continuously extendable over X.

I want to prove that a completely regular space X is an F-space iff any two disjoint sets functionally open in X are completely separated. $(\leftarrow :)$ It is known that any disjoint functionally closed sets A,B in X are completelty separated.Hence, every disjoint functionally open sets are completely separated. So, one side is clear. $(:\rightarrow )$ For the other side , lets take a functionally open set M and take a continuous function $f:M\rightarrow I$. I need a two disjoint functionally open set in X. One of them is M. For the other, I take $N={f^{-1}}(0)$ is functionally closed and complement of N in X is functionally open. I say $N^{c}\cap M=\emptyset$. If it is true, then I can use the completely separated property. then there exists a continous function $h:X\rightarrow I$ such that $h(x)=0$ for each $x\in M$ and $h(x)=1$for each $x\in N^{c}$ Hence, can we say h is the extension of f over X?

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Sorry, what is 'functionally open'? And exactly which point are you uncertain? –  Berci Nov 20 '12 at 9:49
    
@Berci: A functionally open set is what some of us know as a cozero-set: it’s the inverse image of an open set under some continuous real-valued function $f:X\to\Bbb R$. –  Brian M. Scott Nov 20 '12 at 10:14
    
No, you can’t say that $N^c\cap M=\varnothing$: that’s true only if $M\subseteq f^{-1}[\{0\}]$, which is true iff $f$ is the zero function on $M$. –  Brian M. Scott Nov 20 '12 at 10:29
    
you are right.. So, could you give me any hint? –  ege Nov 20 '12 at 10:35
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@M.Sina: No: $U=\Bbb R\setminus\Bbb Z$ is functionally open, and the floor function is continuous on $U$, but it can’t be extended continuously to $\Bbb R$. –  Brian M. Scott Feb 8 '13 at 4:57

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