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I read the proof of the Nested Interval Property for bounded closed intervals ,so I think it's also true for bounded half closed intervals.

Nested Interval property for bounded half closed intervals: $\forall n \in\mathbb{N}$, assume we are given a bounded half closed interval $I_{n}=(a_{n},b_{n}]$.Assume that each $I_{n}$ contains $I_{n+1}$.Then, the resulting nested sequence of closed intervals

$I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \cdots $ has a nonempty intersection; that is, $$ \bigcap_{n=1}^{\infty}I_{n} \neq \emptyset $$

I think the same proof works for it.

Mahmoud albahar
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1 Answers1

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My try to prove it in similar way:

Proof: In order to show that $\bigcap_{n=1}^{\infty} I_{n}$ is not empty, we are going to use the Axiom of Completeness (AoC) to produce a single real number x satisfying $x \in I_{n} \forall n \in \mathbb{N}$.Consider the set $$ B=\{b_{n}:n \in \mathbb{N} \} $$

of right-hand endpoints of the intervals. Because the intervals are nested, we see that every $a_{n}$ serves as an lower bound for B. Thus, we are justified in setting $$x=infA$$ Now, consider a particular $I_{n}=(a_{n},b_{n}]$. Because $x$ is an lower bound for B, we have $x \leq b_{n} $. The fact that each $a_{n}$ is a lower bound for $B$ and that $x$ is the greatest lower bound implies $a_{n}< x$. Altogether then, we have $a_{n} < x \leq b_{n}$,which means $x \in I_{n} \forall n \in \mathbb{N}$.

Note: It looks that the proof would be false if $x \in A=\{a_{n}:n \in \mathbb{N} \}$ but this won't be the case since $a_{1}<a_{2}<a_{3}< \cdots$ and $A$ has infinite number of elements.

But it would be false if $A$ contains finite number of elements as in

https://math.stackexchange.com/a/1370698/589

we have $A=\{0\}$.

Mahmoud albahar
  • 1,083