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$X= \bigcup K_m$, where $K_m$ is a increasing sequence of compact sets and $X$ is the open set.

Aloizio Macedo
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INDU BARAN MANDAL
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  • What background results do you know? If you can give a full characterization of the open sets in $\mathbb{R}^n$, this is a rather short argument. If not, then there are some other arguments that apply the properties of metric spaces. What have you tried? –  Jan 20 '16 at 22:49
  • If you are unsure where to begin, start with the easy special cases. Here I would consider the basic open sets of $\mathbb{R}$, which are open intervals. How would you write one of these as a nested union of compact sets? – hardmath Jan 20 '16 at 23:02
  • i know basic topology @T.Bongers – INDU BARAN MANDAL Jan 21 '16 at 12:18

2 Answers2

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Every open set in $\mathbb{R}^k$ is a increasing union of closed sets. Namely, if the open set is $A$, take $C_n:=\{x \mid d(x,A^c) \geq \frac{1}{n}\}.$ Consider now the sequence $K_n=[-n,n]^k$. We then have that the sequence $K_n \cap C_n$ is what you need.

Note that the argument holds for any $\sigma$-compact metric space, and the fact that an open set is the union of a increasing sequence of closed sets holds in any metric space.

Aloizio Macedo
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    K_n should be K_n ={ x | ||x|| <= n } – INDU BARAN MANDAL Jan 21 '16 at 12:21
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    Can you also prove that $C_n$ is convex if A is convex? It seems to be true in $R^n$ with the Euclidean norm but I can't prove it – Noel May 11 '20 at 07:11
  • It's would be proved if you are able to prove that $d(x,A^{c})=f(x)$ is a concave function. – R. W. Prado Mar 18 '21 at 19:45
  • The sets $K_n \cup C_n$ only cover the interior of $A.$ There is also a boundary of $A$ which is a non-empty subset of $\mathbb R^k$ (unless $A = \varnothing$ or $\mathbb R^k$) and the distance of $A^c$ from the boundary points of $A$ is zero. So this forces us to take $\left {K_n \cap (C_n \cup \partial A) \right }_{n \geq 1}$ to be the desired increasing sequence of compact sets whose union is $A.$ – Akiro Kurosawa Aug 26 '23 at 20:34
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Let's start with an open interval $(a,b)$. If $a=-\infty$ xor $b=\infty$ then we have $$(a,b)=\bigcup\limits_{n=1}^\infty [b-n,b-1/n]$$ or $$(a,b)=\bigcup\limits_{n=1}^\infty [a+1/n,a+n]$$ if both are infinite then $$(a,b)=\bigcup\limits_{n=1}^\infty [-n,n]$$ and if neither of them are, let $c=\frac{b-a}4$ then $$(a,b)=\bigcup\limits_{n=1}^\infty [a+c/n,b-c/n]$$ so any open interval can be written as the union of an increasing sequence of compact sets. From here, recall that any open set $O$ is a disjoint union of countably many open intervals, say $$O=\bigcup_{n=1}^\infty A_n$$ for intervals $A_n$. By our earlier work, for all applicable $n$, $$A_n=\bigcup_{i=1}^\infty C_{n,i}$$ for an increasing sequence of compact sets $\{C_{n,i}\}$. Then for each applicable fixed $i$ let $$C_i=\bigcup_{n=1}^\infty C_{n,i}.$$ Then $\{C_i\}$ is an increasing sequence of compact sets with $$O=\bigcup_{i=1}^\infty C_i.$$

Sean English
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