$X= \bigcup K_m$, where $K_m$ is a increasing sequence of compact sets and $X$ is the open set.
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put on hold as off-topic by Jendrik Stelzner, Daniel Robert-Nicoud, Norma, Rory Daulton, Pragabhava 2 days agoThis question appears to be off-topic. The users who voted to close gave this specific reason:
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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. |
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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 $$O=\bigcup_{i=1}^\infty C_i.$$ |
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