Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ be a subset of first category of product space $X \times Y$. Why is the following true: if $(\bar E)_x \subset Y$ is of first category then it follows that $E_x$ is nowhere dense. $E_x$ denotes the $x$-section of E: the projection of $E \cap (\{x\} \times Y)$ down to $Y$.

share|cite|improve this question
If by $(\bar E)_x$ you mean the topological closure of $E_x$, then what you're asking has nothing to do with product spaces and (presumably, although you didn't mention it) the Kuratowski-Ulam Theorem. I don't know how general you want "space" to be (and it is likely this will involve technical details that I don't know much about, such as Baire spaces and Cech complete spaces and the like), but it's pretty trivial that "closed and first category" implies "nowhere dense" in ${\mathbb R}^n.$ In fact, even in a complete metric space, "$G_{\delta}$ and first category" implies "nowhere dense." – Dave L. Renfro Nov 28 '11 at 17:23
I see now from the title (but not anywhere in the question itself) that $(\bar E)_x$ is intended to be the $x$-section of the closure of $E.$ The projection of a closed set does not have to be closed (e.g., the projection onto the $x$-axis of the graph of $xy = 1$ in ${\mathbb R}^2$), so I'm not sure where to go off-hand. I still think you need some assumptions on the spaces $X$ and $Y,$ however. – Dave L. Renfro Nov 28 '11 at 17:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.