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Let $A \subset \mathbb{R}^n$ be a compact set, $\epsilon \in \mathbb{R}_{>0}$, $O := A + \epsilon \mathbb{B}^\circ$, and define the open set

$$ \bar O := O \times \mathbb{R}^m. $$

Let $C \subseteq \mathbb{R}^n$ be a closed set such that $C \supset O$.

Consider a closed set $\bar C \subseteq \mathbb{R}^n \times \mathbb{R}^m$ such that its projection to $\mathbb{R}^n$ is $C$, i.e. $C$ is the maximal set such that $\forall x\in C \ $ $\exists y \in \mathbb{R}^m$ such that $(x,y) \in \bar C$.

Question: is the set $S := \bar O \cap \bar C$ open?

Comment: as $O \subset C$ we have $\bar C \nsubseteq \bar O$. Seems that the intersection of an open set, $\bar O$, with a closed one, $\bar C$, which is not a subset, should be open as well.

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What topology do you put on the whole space? – Rudy the Reindeer Sep 19 '12 at 9:05
@Matt: That’s understood: it’s the Euclidean topology. – Brian M. Scott Sep 19 '12 at 9:07
Yes, Euclidean. But $O$ is particular, otherwise it does not work. Say $O = A + \epsilon \mathbb{B}^\circ$, with $A$ compact. – Adam Sep 19 '12 at 9:08
$C\times\varnothing=\varnothing$, so it won’t tell you much. – Brian M. Scott Sep 19 '12 at 9:18
Thanks. Updated. – Adam Sep 19 '12 at 9:21
up vote 1 down vote accepted

Added: Despite the significant change in the hypothesis on $O$, this hint still applies.

HINT: Let $z\in\Bbb R^m$ be the point $\langle 0,\dots,0\rangle$. Show that $C\times\{z\}$ satisfies the conditions imposed on $\bar C$. For this particular $\bar C$, what is $S$? Is it open in $\Bbb R^{n+m}$?

You might find it very helpful to draw a picture of the case $n=m=1$.

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I made small examples. Seems to me that if $\bar O$ has no boundary, say $O = A + \epsilon \mathbb{B}^\circ$, then $S$ is open as well because $\bar O \nsupseteq \bar C$. – Adam Sep 19 '12 at 9:32
@Adam: Look at my example. For it $S$ is definitely not open. – Brian M. Scott Sep 19 '12 at 9:34
What is $O$ in your example? – Adam Sep 19 '12 at 9:36
@Adam: It doesn’t matter. No matter what open set $O$ is, $S$ is not open. – Brian M. Scott Sep 19 '12 at 9:38
@Adam: That shows that $S$ is not closed. It’s also not open. Not closed does not imply open: a set can be both open and closed, open but not closed, closed but not open, or neither open nor closed. – Brian M. Scott Sep 19 '12 at 9:44

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