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Let $K\subset \mathbb{R}^N$ be a compact set. We say $K$ is "good" if the following property holds:

Given a set of open neighborhoods $\{x\in U_x\subset \mathbb{R}^N\}_{x\in K}$ there exists a finite set $S\subset K$ and relatively compact open subsets $x\in \tilde{U}_x\subset U_x$ such that

  1. $\{\tilde{U}_x\}_{x\in S}$ covers K,

  2. For all $x,y\in S$, $\tilde{U}_x\cap \tilde{U}_y\neq \emptyset$ implies $\tilde{U}_x\cap \tilde{U}_y \cap K \neq \emptyset$.

The question is: Can a compact set be ``bad"?

While this may sound obvious, I am worried about crazy looking cantor type compact sets.

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  • $\begingroup$ (Assuming that by "bad" you mean the negation of "good"?) $\endgroup$ – user62017 Jan 28 '15 at 15:41
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    $\begingroup$ @Goldstern, I believe the OP means relatively compact in the sense that the closure of $\bar U_x$ in the relative topology of $U_x$ is compact (that is, $U_x$ contains the closure of $\bar U_x$). $\endgroup$ – Joonas Ilmavirta Jan 28 '15 at 20:16
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    $\begingroup$ N is just a number. $\endgroup$ – Mohammad F. Tehrani Jan 28 '15 at 21:42
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    $\begingroup$ Let's say we just use balls as our $\widetilde{U}$'s. If two balls intersect, then either: (1) the boundary of the intersection does not intersect $K$: replace the two balls by 3 disjoint open sets; or (2) the boundary has points of $K$ in it: make one of the balls a little larger, and again I'm good. $\endgroup$ – Christian Remling Jan 28 '15 at 23:19
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    $\begingroup$ @ChristianRemling: I don´t quite see how you handle the interference of fixing two different pairs of balls. $\endgroup$ – Ramiro de la Vega Jan 29 '15 at 13:20
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This is an attempt to formalize Christian Remling´s idea.

Cover $K$ by finitely many balls $\{B(x,r_x): x \in S \}$ such that $\overline{B(x,2r_x)} \subseteq U_x$. Let $\delta$ be the minimum of the $r_x$'s, and for each pair $x,y \in S$, find a positive real $\lambda_{x,y}<\delta$ such that: $$\overline{B(x,r_x) \cap B(y,r_y)} \cap K=\emptyset \implies \overline{B(x,r_x+\lambda_{x,y}) \cap B(y,r_y+\lambda_{x,y})} \cap K=\emptyset.$$ Let $\lambda$ be the minimum of the $\lambda_{x,y}$'s and let $C$ be the union of all the sets $\overline{B(x,r_x+\lambda) \cap B(y,r_y+\lambda)}$ for which $\overline{B(x,r_x) \cap B(y,r_y)} \cap K=\emptyset$. We can now let $$\tilde{U}_x= B(x,r_x+\lambda) \setminus C$$ for each $x \in S$.

Clearly $\{\tilde{U}_x\}_{x\in S}$ still covers $K$ because $C \cap K=\emptyset$. Also $\tilde{U}_x\subset U_x$ because $\lambda < \delta$. Finally, if $\tilde{U}_x\cap \tilde{U}_y\neq \emptyset$, then $B(x,r_x+\lambda) \cap B(y,r_y+\lambda)$ is not contained in $C$. This implies that $\overline{B(x,r_x) \cap B(y,r_y)} \cap K \neq \emptyset$ and therefore $\tilde{U}_x\cap \tilde{U}_y \cap K\neq \emptyset$.

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  • $\begingroup$ This proof readily extends to compact sets inside any metrizable topological space, instead of $\mathbb{R}^n$. Do you see a way of changing the proof that does not involve the use of metric. $\endgroup$ – Mohammad F. Tehrani Jan 31 '15 at 0:58

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