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Let $\{[a_n,b_n]\}_{n\in\mathbb{N}}$ be a sequence of closed bounded intervals in $\mathbb{R}$ such that $(\forall n\in\mathbb{N})([a_{n+1},b_{n+1}]\subset [a_n,b_n])$ e $\lim_{n\to\infty}(b_n-a_n)=0$. Let $S\subset [a_0,b_0]$ a set such that $(\forall n\in\mathbb{N})(S\cap [a_n,b_n]\neq \emptyset )$.

We know from the Nested Intervals Theorem that exists $\sigma\in\mathbb{R}$ such that $\bigcap _{n=1}^\infty [a_n,b_n]=\{\sigma\}$. My question: the element $\sigma$ belongs to the set $S$? That is, is it true that $\sigma\in S$?

I have not been able to prove either that it is false or that it is true. I tried to prove that $\sigma\in S$ by contradiction:

Assume that $\sigma\notin S$, then $S\cap \left(\bigcap _{n=1}^\infty [a_n,b_n]\right)=\emptyset $. This implies that $(\forall x\in S)\left(x\notin \bigcap _{n=1}^\infty [a_n,b_n]\right)$.

Given $x\in S$, we have $x\notin \bigcap _{n=1}^\infty [a_n,b_n]\Rightarrow (\exists n\in\mathbb{N})(x\notin [a_n,b_n])$.

Therefore,

$(\forall x\in S)\left(x\notin \bigcap _{n=1}^\infty [a_n,b_n]\right)\Rightarrow (\forall x\in S)(\exists n\in\mathbb{N})(x\notin [a_n,b_n])$

But I could think of nothing else. Since $(\forall x\in S)\left(x\notin \bigcap _{n=1}^\infty [a_n,b_n]\right)$ is true, we have that $ (\forall x\in S)(\exists n\in\mathbb{N})(x\notin [a_n,b_n])$ is also true. But I can not infer from this proposition that $\sigma\in S$.

Can someone help me please?

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  • $\begingroup$ Is there a material difference between $a_n$ and $a^{(n)}$? $\endgroup$ – Saucy O'Path Sep 10 '18 at 21:48
  • $\begingroup$ No, I just forgot to change the notation. I will correct. $\endgroup$ – user477271 Sep 10 '18 at 21:55
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By the limit condition, at least one of the sequences $(a_n)$ and $(b_n)$ is not eventually constant, say $(a_n)$. Then if you take $S=\{a_n\mid n\in\mathbb N\}$, $\sigma$ will not belong to $S$. So, the answer is: $\sigma$ doesn't have to belong to $S$.

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  • $\begingroup$ I have noticed that in fact $\sigma\in S$ in the case where $S$ is a closed set. $\endgroup$ – user477271 Sep 10 '18 at 22:41
  • $\begingroup$ To the proposer: Suppose $a_n<a_{n+1}<b_{n+1}<b_n$ for every $n$ and $S_1=[a_0,b_0] $ while $S_2=[a_0,b_0]\backslash \{\sigma\}.$ $\endgroup$ – DanielWainfleet Sep 11 '18 at 6:53

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