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How can i prove it?

[Gillman and Jerison] If a dense subspace $Y$ of a Tychonoff space $X$ is $C-embedded$ in X, then $Y$ is $ G‎‎_{\delta‎‎‎}-dense‎ $‎ in $X$.

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$\newcommand{\cl}{\operatorname{cl}}$Suppose that $G$ is a non-empty $G_\delta$ in $X$ such that $G\cap Y=\varnothing$. Then there are open sets $U_n$ for $n\in\Bbb N$ such that $G=\bigcap_{n\in\Bbb N}U_n$ and $U_n\supseteq U_{n+1}$ for each $n\in\Bbb N$. Fix $x\in G$. Since $X$ is Tikhonov, for each $n\in\Bbb N$ there is a continuous function $f_n:X\to[0,1]$ such that $f_n(x)=1$ and $f_n(y)=0$ for all $y\in X\setminus U_n$. Now define

$$f:Y\to\Bbb R:y\mapsto\sum_{n\in\Bbb N}f_n(y)\;;$$

each $y\in Y$ belongs to only finitely many of the sets $U_n$, so $f$ is well-defined. Does $f$ have a continuous extension to $X$?

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    Dense subspace $Y$ of $X$ does it matter?2012-11-18
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    @MohammaDSina: If $Y$ weren’t dense in $X$, it couldn’t possibly be $G_\delta$-dense, so in that sense it matters. It does not matter for answering the question that I left at the end, however. To answer that you need only look carefully at the definition of $f$.2012-11-18
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    Why $f$ does not have a countinuous extension on $X$?2013-03-07
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    @M.Sina: Let $V$ be any open nbhd of $x$. For $k\in\Bbb N$ let $W_k=\{y\in U_k:f_k(y)>1/2\}$, and for $n\in\Bbb N$ let $V_n=V\cap\bigcap_{k\le n}W_k$. Then $V_n$ is an open nbhd of $x$ contained in $V$, and $f(y)>\frac{n}2$ for each $y\in V_n\cap Y$. Thus, $x$ has arbitrarily small nbhds on which $f$ is arbitrarily large, and it’s impossible to extend $f$ continuously $X$.2013-03-07
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    Thank you for following this discussion. In [Topological Groups and Related Structures](http://books.google.com/books?id=B785AETmFKEC&pg=PA347&dq=Theorem+6.1.4+L.+gillman+and+M.+jerison&hl=en&sa=X&ei=lvxDUYvHD-mZiAeJ_oHYAg&ved=0CC0Q6AEwAA#v=onepage&q=Theorem%206.1.4%20L.%20gillman%20and%20M.%20jerison&f=false) Theorem 6.1.4. Is this proof is correct? $P \subset X\setminus Y$ and why $g$ cannot be continuously extend over $X$?2013-03-16
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    @M.Sina: Yes, that proof is fine. Simpler than mine, too. To see why $g$ can’t be extended, fix $p\in P$. $Y$ is dense in $X$, so $p\in\operatorname{cl}Y$. For each $\epsilon>0$ there is an open nbhd $U_\epsilon$ of $p$ such that $f(x)<\epsilon$ for all $x\in U_\epsilon$, and therefore $g(x)>\frac1{\epsilon}$ for all $x\in U_\epsilon\cap Y$. Since $g(x)$ assumes arbitrarily large values on arbitrary nbhds of $p$, $g$ can have no continuous extension to $p$.2013-03-16
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    In Arhangelskii's proof, he says that since $X$ is tychonoff there exists a non-empty zero-set $P$ in $X$ contained in $X\setminus Y$. Why tychonoffness implies existence of $P$ as a zero-set? ( he assume that $Y$ is not $G_\delta$-dense in $X$ and therefore $P$ is a $G_\delta$ set in $X$ contained in $X\setminus Y$ ).2013-09-10
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    @M.Sina: It’s a general fact that zero sets are closed under countable intersection. (If $Z_n$ is the zero set of $f_n$ for $n\in\Bbb N$, let $g_n=\min\{f,2^{-n}\}$ and $g=\sum_{n\in\Bbb N}g_n$. The series converges uniformly, so $g$ is continuous, and its zero set is $\bigcap_n Z_n$.) Now let $G=\bigcap_nU_n\subseteq X\setminus Y$ be a $G_\delta$, and fix $x\in G$. $X$ is Tikhonov, so for each $n\in\Bbb N$ there is a zero set $Z_n$ such that $x\in Z_n\subseteq U_n$. Let $P=\bigcap_n Z_n$.2013-09-10
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    @M.Sina: You’re welcome.2013-09-11