# Is the nested interval property true for bounded half closed interval?

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.

• Well, try, and pay attention. Commented Apr 9, 2023 at 19:44
• The proof for closed intervals is not long. If you "think the same proof works" for half closed intervals, you should reproduce the proof in your question, so someone here could find a flaw. Commented Apr 9, 2023 at 23:43
• – lhf
Commented Apr 10, 2023 at 3:04
• @lhf I am asking about bounded half closed interval not open intervals. Commented Apr 10, 2023 at 3:15
• Well, what went wrong with open intervals probably goes wrong with half-open intervals. Commented Apr 10, 2023 at 4:13

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 $$a_{1} and $$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 $$A=\{0\}$$.