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In Royden's Real Analysis P164 Q7.42b, It assumes that $X$ and $Y$ are complete metric spaces. Let $O$ be a dense open set in $X \times Y$. Assertion: Then there is a $G \subset X$ which is a dense $G_\delta$ such that $E_x = \{ y \in Y : (x,y) \in O \}$ is a dense open subset of $Y$ for each $x \in G$.

My question is: I think there is a missing condition in the question. For general $Y$, I am not sure that this holds. However, if we assume $Y$ is second countable, then the assertion holds.

I think this is just a version of the Kuratowski-Ulam Theorem. In the Oxtoby book, there is some requirements on $Y$.


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There's a pretty substantial list of errata on the new coauthor's site (Fitzpatrick, I think the name is?). Did you check there? – Dylan Moreland May 13 '12 at 19:59
Errata for the 1st printing of the 4th ed. are here; this appears to be a different edition from yours, but I didn’t see anything obviously corresponding to this question. – Brian M. Scott May 13 '12 at 20:11
Yes. I have already checked the 4th edition. This question is not there. – XxXxX May 13 '12 at 20:27
up vote 2 down vote accepted

Yes, a hypothesis is definitely missing. The usual proof requires that $Y$ have a countable $\pi$-base, i.e., a countable family $\mathscr{C}$ of non-empty open sets such that every non-empty open subset of $Y$ contains a member of $\mathscr{C}$; here’s an example showing that the result can fail in the absence of such a hypothesis.

Let $Y=[0,1]\times[0,1]$, and define a metric $d$ on $Y$ as follows:

$$d\Big(\langle x,a\rangle,\langle y,b\rangle\Big)=\begin{cases} |x-y|,&\text{if }a=b\\ 2,&\text{if }a\ne b\;. \end{cases}$$

It’s easy to check that $d$ is a complete metric. It will be convenient to let $Y_a=\{\langle x,a\rangle:x\in[0,1]\}$ for each $a\in[0,1]$; note that $Y_a$ is open in $Y$.

Let $X=[0,1]$ with the usual topology; $X\times Y$ is the disjoint union of the clopen subsets $X\times Y_a$ for $a\in[0,1]$, and each of these subsets is homeomorphic to $[0,1]\times[0,1]$ with the usual topology.

Now let

$$O=\left\{\Big\langle x,\langle y,a\rangle\Big\rangle\in X\times Y:x\ne a\right\}\;.$$

For each $a\in[0,1]$, $O\cap\Big(X\times Y_a\Big)$ is homeomorphic to $\Big([0,1]\setminus\{a\}\Big)\times[0,1]$ with the usual topology and therefore is dense and open in $X\times Y$, so $O$ itself is dense and open in $X\times Y$. But for each $x\in X$, $E_x=\{\langle y,a\rangle\in Y:a\ne x\}$, which is disjoint from $Y_x$ and therefore not dense in $Y$. That is, $\{x\in X:E_x\text{ is dense and open in }Y\}=\varnothing$.

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Thanks for your answer. Here is a slightly simplified example: Let $Y$ be $[0,1]$ with the discrete topology, i.e. $d(x,y) = 2$ if $x \ne y$, and let $X$ be $[0,1]$ with the usual metric. Then $O = \{(x,y) | x \ne y \}$ is an open and dense set in $X \times Y$. Hence, for any $x \in X$, $ E_x = \{y \in Y | y \ne x \} $ is an open and closed set in $Y$. – XxXxX May 17 '12 at 12:24

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