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So I'm looking at this proof, which is presented as a problem in Gamelin & Greene but I'm having some trouble understanding it.

Part (a), (b) and (c) are straightforward enough. However I'm having trouble following his reasoning on (d). I understood how everything is defined, I just don't see how the Baire Category Theorem applies, which says that the intersection of an infinite sequence of dense subsets of a complete metric space is dense. Could someone please explain it further?

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Can I ask what is the $J_\delta(t_m)$ in d)? – Hui Yu Oct 9 '12 at 15:53
You believe that the Baire Category Theorem applies to $\mathbb{R}$, right? And that a set dense in the Sorgenfrey topology is also dense in the regular topology? – rschwieb Oct 9 '12 at 16:33

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$We have an open set $V$ in $\Bbb R_S^2$ that contains $F$, where $F=\{\langle x,-x\rangle:x\in T\}$, and $T$ is $\Bbb R\setminus S$ for some countable dense subset (not ‘sequence’: van Gool’s terminology is wrong) of $\Bbb R$ with respect to the usual Euclidean topology. For any $x\in\Bbb R$ and $\epsilon>0$ we define $J_\epsilon(x)=[x,x+\epsilon)\times[-x,-x+\epsilon)$, a basic open nbhd of $\langle x,-x\rangle$ in $\Bbb R_S^2$.

For $n\in\Bbb Z^+$ let $T_n=\{t\in T:J_{1/n}(t)\subseteq V\}$; I claim that $\bigcup_{n\in\Bbb Z^+}T_n=T$.

Proof. Since $V$ is an open nbhd of $F$, for each $t\in T$ there is an open nbhd of $\langle t,-t\rangle$ contained in $V$. Thus, for each $t\in T$ there is some $\epsilon(t)>0$ such that $\langle t,-t\rangle\in J_{\epsilon(t)}(t)\subseteq V$. Then for any positive integer $n>\frac1{\epsilon(t)}$ we have $\frac1n<\epsilon(t)$ and therefore $\langle t,-t\rangle\in J_{1/n}(t)\subseteq J_{\epsilon(t)}(t)\subseteq V$, i.e., $t\in T_n$. $\dashv$

Suppose that $\int\cl T_n=\varnothing$ for each $n\in\Bbb Z^+$, where the interior and closure are taken with respect to the usual Euclidean topology on $\Bbb R$. For $n\in\Bbb Z^+$ let $G_n=\Bbb R\setminus\cl T_n$; then each $G_n$ is a dense open subset of $\Bbb R$. For each $s\in S$ let $G_s=\Bbb R\setminus\{s\}$; $G_s$ is clearly also a dense open subset of $\Bbb R$. Let $$G=\bigcap_{n\in\Bbb Z^+}G_n\cap\bigcap_{s\in S}G_s\;;$$ $S$ is countable, so by the Baire category theorem $G$ is dense in $\Bbb R$. In particular, $G\ne\varnothing$, so

$$\begin{align*}\Bbb R&=S\cup T\subseteq\bigcup_{s\in S}\{s\}\cup\bigcup_{n\in\Bbb Z^+}\cl T_n\\ &=\bigcup_{s\in S}(\Bbb R\setminus G_s)\cup\bigcup_{n\in\Bbb Z^+}(\Bbb R\setminus G_n)\\ &=\Bbb R\setminus\left(\bigcap_{s\in S}G_s\cap\bigcap_{n\in\Bbb Z^+}G_n\right)\\ &=\Bbb R\setminus G\\ &\ne\Bbb R\;, \end{align*}$$

which is absurd. Thus, there must be an $n\in\Bbb Z^+$ such that $\int\cl T_n\ne\varnothing$.

Note that this argument works generally, not just in this setting: a space $X$ has the property that every intersection of countably many dense open sets is non-empty iff it is not the union of countably many nowhere dense closed sets.

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