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I learn real analysis and topology then I found something interesting about constant function. I am unsure it is true or false because I cannot prove it. I found property as follows:

If $X$ is $T_1$ and every continuous function $f:X\rightarrow \mathbb{R}$ is constant then for any proper nonempty closed set $F$, $\bigcap\{U(x): U \mbox{ is open set containing } x, \mbox{ for any } x\in F\}\neq\emptyset.$

For instance $X$ is uncountable space equipped with co-countable topology. Then every proper nonempty closed set $F$ of $X$ satisfies above condition.

Please, help me to prove it or give me counterexample if it is false.

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    $\begingroup$ The definition of $\{ U(x)| U \mbox{ is open set containing $x$, for any $x\in F$} \}$ is not clear. Do you mean that it is the set of open sets intersecting $F$ in some point $x \in F$? $\endgroup$
    – Crostul
    Nov 29, 2014 at 18:57
  • $\begingroup$ Yes, it is. But the intersection of sets $\{U(x):U \mbox{ is open set containing } x, \mbox{ for any } x\in F\}$ might be not subset of $F$ $\endgroup$
    – narita fung
    Nov 30, 2014 at 1:58
  • $\begingroup$ It would be clearer to say $\bigcap\{U: U \mbox{ is open and } U \cap F \not= \emptyset \}\neq\emptyset$ if that is what you mean. $\endgroup$
    – Rob Arthan
    Nov 30, 2014 at 11:07
  • $\begingroup$ @Rob Arthan. Yes,that I mean $\endgroup$
    – narita fung
    Nov 30, 2014 at 11:31
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    $\begingroup$ The cocountable example does not seem to work. If $X$ is $\mathbb{N}$ with the cocountable topology, and $f: \mathbb{N} \to \mathbb{R}$, $n \mapsto n \operatorname{mod} 2$, then $f$ is continuous but not constant. $\endgroup$
    – Hew Wolff
    Nov 30, 2014 at 19:16

1 Answer 1

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Your example of an uncountable set $X$ equipped with the cocountable topology provides a counterexample.

Any continuous function $X \rightarrow \mathbb{R}$ is constant. To see this, assume $f$ is a continuous function $X \rightarrow \mathbb{R}$ and that $x, y \in X$ have $a = f(x) < b = f(y)$. $X$ is connected under the cocountable topology (since any two non-empty open sets meet) and so the image of $f$ must contain the closed interval $[a, b]$. But then $f^{-1}([a, (a+b)/2])$ is an uncountable closed subset of $X$ and hence must be $X$ itself, which is impossible because $y \not\in f^{-1}([a, (a+b)/2])$. So, according to the conjecture, if $F$ is any non-empty closed subset of $X$, the intersection of all the open sets that meet $F$ is non-empty. But this is not so: if $F$ is a closed subset of $X$ with more than one element, and $x \in X$, let $U = X \mathop{\backslash} \{x\}$; then $U$ is open, $U$ meets $F$ (because $F$ has more than one element) and $x \not\in U$.

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  • $\begingroup$ For every open set $U$ with $U\cap F\neq\emptyset$ we have $X-U$ is countable and $\bigcup X-U$ is countable. Therefore $\bigcap U\neq\emptyset$. $\endgroup$
    – narita fung
    Dec 1, 2014 at 9:19
  • $\begingroup$ I am not sure what you mean by $\bigcup X - U$. If you mean the union of the sets $X - U$ as $U$ ranges over open sets with $U \cap F \not= \emptyset$, then that is the union of uncountably many countable sets, and need not itself be countable. $\endgroup$
    – Rob Arthan
    Dec 1, 2014 at 13:35

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