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Show that there is a sequence $\{K_n\}$ of compact subsets such that $ K_1 \subset K_2 \subset K_2^\circ \subset K_3 \subset K_3 ^\circ \subset \cdots$ such that the nonempty open subset in $\mathbb{C}$, $O$, $O = \bigcup_{n=1}^{\infty} K_n$

My Attempt:

Let $A =\{x \in O : d(x,O^c) \le r\}$ where $A$ is closed and bounded and therefore compact. Now let $K_n=\{x \in O: d(x,O^c)=r/n\}$ and each $K_n$ is compact as it is a collection of points.

Notice that each compact set $K_i$ will "collect" each $x \in O$. Furthermore, since $O$ is open the compact subset $K_n$ as $n \rightarrow \infty$ will fail to converge since there exist no $x \in O$ such that $d(x, 0^c)=0$ and so $O = \bigcup_{n=1}^\infty K_n$

I think my use of "collect" displays the lack of rigor (or even validity) of my proof. Any advice is greatly appreciated.

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  • $\begingroup$ My universe is $\mathbb{C}$. I will edit and include this. $\endgroup$ – user7090 Nov 8 '13 at 3:19
  • $\begingroup$ Am I understanding your question wrongly, or do you want, given an open set $O$, find compact sets $K_n$ with $K_1 \subset K_2^\circ\subset K_2 \subset K_3^\circ \subset K_3 \subset K_4 ^\circ \subset \cdots$ (note you have something different) such that $\bigcup K_n=O$? $\endgroup$ – Pedro Tamaroff Nov 8 '13 at 3:41
  • $\begingroup$ @Pedro I am given an open set $O$ of which I need to find compact sets $K_n$. $\endgroup$ – user7090 Nov 8 '13 at 3:45
  • $\begingroup$ Yes, that is what I thought. $\endgroup$ – Pedro Tamaroff Nov 8 '13 at 3:45
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Let's do it in a very specific metric space; the idea can be easily generalized.

Consider $\Bbb{R}$ and sets of the form

$$K_n = \left[\frac{1}{n}, 1 - \frac{1}{n}\right]$$

Suppose that $x \in (0, 1)$; by the Archimedian property, we can choose an $N$ large enough that $$\frac{1}{N} < \min \left(x, 1 - x\right)$$

Hence $x \in K_n$. It's immediate to see that any element not in $(0, 1)$ cannot be in any $K_n$, so the union is simply

$$\bigcup_n K_n = (0, 1)$$ which is open.


More generally, try considering all the elements whose distance from a given fixed point is at most $n$, but the distance from the boundary of the desired $O$ is at least $\frac{1}{n}$.

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It's known that $O$ can be writen as countable union of closed balls : $$O=\mathop\cup_{n \in {\mathbb N}}B_n$$ Let us put : $$K_n=\mathop\cup_{k=0}^n B_k$$

It's clear thet $K_n$ are compacts such tahte $\forall n \in {\mathbb N}, K_n \subset K_{n+1}$and : $\mathop \cup_{n\in {\mathbb N}} K_n = O$

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Let $O\subseteq \Bbb R^2$ be your open set. Consider the (compact) sets $$K_n=\{{\bf x}=(x,y):{\bf x}\in O, d({\bf x},{\bf 0})\leqslant n,d({\bf x},\partial O)\geqslant n^{-1}\}$$

ADD $\partial O$ denotes the boundary of $O$.

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    $\begingroup$ $\bigcup_{n=1}^{\infty}K_n$ is supposed to equal $O$. Suggest you change $\partial O$ to $\mathbb R^2\setminus O$. $\endgroup$ – bof Nov 8 '13 at 4:36
  • $\begingroup$ @bof Shouldn't $\bf{x}$ be in $O$ so that $K_n\subset O$ ? $\endgroup$ – Hua Apr 25 '17 at 12:59

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