Proof of Tail Suprema and Tail Infima of a Bounded Sequence of Real Numbers

Each result cited is universally quantified over the data in its own statement.

Clause tails. By Bounded Sequences of Real Numbers §bounded there is $M\in\mathbb{R}$ with $|a_{n}|\le M$ for every $n\in\mathbb{N}$, so $-M\le a_{n}\le M$ for every $n\in\mathbb{N}$ by the ordered-field rules. Fix $k\in\mathbb{N}$. Then $M$ is an upper bound and $-M$ a lower bound of $T_{k}$ as in Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds, so $T_{k}$ is bounded by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded; and $a_{k}\in T_{k}$ because $k\le k$ by Arithmetic and Order of the Natural Numbers §partial-order, so $T_{k}$ is nonempty. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum and The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §infimum, $T_{k}$ has a supremum and an infimum, each unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum; so $\overline{a}_{k}=\sup T_{k}$ and $\underline{a}_{k}=\inf T_{k}$ are defined.

Clause sandwich. Let $k\in\mathbb{N}$. By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least, $\overline{a}_{k}$ is an upper bound of $T_{k}$ that is $\le$ every upper bound of $T_{k}$, and $\underline{a}_{k}$ is a lower bound of $T_{k}$ that is $\ge$ every lower bound of $T_{k}$; call this (E). As $a_{k}\in T_{k}$ by the proof of clause tails above, (E) gives $\underline{a}_{k}\le a_{k}\le\overline{a}_{k}$.

Clause monotone. Let $M$ be as in the proof of clause tails and let $k\in\mathbb{N}$. By clause sandwich above, $-M\le a_{k}\le\overline{a}_{k}$ and $\underline{a}_{k}\le a_{k}\le M$. Every $m\in\mathbb{N}$ with $m\ge k+1$ satisfies $m\ge k$, because $k<k+1$ by Arithmetic and Order of the Natural Numbers §successor and the order is transitive by Arithmetic and Order of the Natural Numbers §partial-order; so $T_{k+1}\subseteq T_{k}$. Hence $\overline{a}_{k}$ is an upper bound and $\underline{a}_{k}$ a lower bound of $T_{k+1}$, and (E) for $T_{k+1}$ gives $\overline{a}_{k+1}\le\overline{a}_{k}$ and $\underline{a}_{k}\le\underline{a}_{k+1}$. As $k$ was arbitrary, $(\overline{a}_{k})$ is nonincreasing and $(\underline{a}_{k})$ is nondecreasing by Monotone Sequences §monotone; and $-M$ is a lower bound of $\{\overline{a}_{k}:k\in\mathbb{N}\}$ and $M$ an upper bound of $\{\underline{a}_{k}:k\in\mathbb{N}\}$, so by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded and Bounded Sequences of Real Numbers §bounded the sequence $(\overline{a}_{k})$ is bounded below and $(\underline{a}_{k})$ is bounded above.

Clause limits. The set $\{\overline{a}_{k}:k\in\mathbb{N}\}$ is nonempty, as it contains $\overline{a}_{1}$, and it is bounded below by clause monotone above and Bounded Sequences of Real Numbers §bounded; so its infimum exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §infimum. Dually, $\{\underline{a}_{k}:k\in\mathbb{N}\}$ is nonempty and bounded above, so its supremum exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum. By clause monotone above, $(\overline{a}_{k})$ is nonincreasing and bounded below, so $\overline{a}_{k}\to\inf\{\overline{a}_{k}:k\in\mathbb{N}\}$ by the second half of Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences §monotone; and $(\underline{a}_{k})$ is nondecreasing and bounded above, so $\underline{a}_{k}\to\sup\{\underline{a}_{k}:k\in\mathbb{N}\}$ by its first half.