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I'm currently reading Abbot's Understanding Analysis, where in page 19 he introduces the Nested Interval Property and writes "because the intervals are nested, we see that every $b_n$ serves as an upper bound for $A$." I was wondering if this claim could be shown more rigorously, and have provided a proof I attempted. Please let me know if there is any error in understanding, or if the proof is sufficient. Thank you in advance.

Hypothesis: Let $A=\{a_n:n\in\mathbb{N}\}$. For all $n,m\in\mathbb{N}$, where $n>m$, $a_n<b_n$, $a_n<a_m$ and $b_n>b_m$. These are the assumptions I made which are relevant to the proof.

Claim: For all $n\in\mathbb{N}$, $b_n$ is an upper bound to $A$.

Proof: Let $i,j\in\mathbb{N}$. We proceed to show that $a_i<b_j$. If $i>j$, then $b_i\leq b_j$; since $a_i<b_i$, we have $a_i<b_j$. If $i<j$, then $b_j\leq b_i$; since $a_i\leq a_j$ and $a_j<b_j$, we have $a_i<b_j$. If $i=j$, then $a_i=a_j<b_j$. Therefore, $a_i<b_j$.

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  • $\begingroup$ Looks fine. ${}$ $\endgroup$ Commented May 14, 2018 at 18:51

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Notice that the a's are decreasing and the b`s are increasing.
Thus $a_1$ is an upper bound of A.
Since $a_1 \lt b_1$ it follows that $b_1$ and all subsequent
b's are are upper bounds of A.

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