Proof of A Bounded Monotone Sequence of Real Numbers Converges

1. (Nondecreasing case.) Assume $a_n\le a_{n+1}$ for every $n\in\mathbb{N}$ and that $A$ is bounded above. Since $A$ is nonempty and bounded above, $s=\sup A$ exists by the least upper bound property recorded in The Real Line: Standing Notation and Background for Calculus on Intervals (ii).

Step 1: $a_n\le a_m$ whenever $n\le m$. Fix $n\in\mathbb{N}$ and let

$$T=\{m\in\mathbb{N}: m<n\ \text{ or }\ a_n\le a_m\} .$$

The least element $1$ of $\mathbb{N}$ lies in $T$: either $1<n$, or $n=1$ and then $a_n\le a_1$ since $a_1\le a_1$. Suppose $m\in T$. If $m<n$, then $m+1\le n$, so either $m+1<n$, or $m+1=n$ and then $a_n\le a_{m+1}$; in both cases $m+1\in T$. If instead $n\le m$, then $a_n\le a_m$ because $m\in T$, and $a_m\le a_{m+1}$ by hypothesis, so $a_n\le a_{m+1}$ and $m+1\in T$. By the principle of induction, $T=\mathbb{N}$. In particular, if $n\le m$ then $m$ is not $<n$, so $a_n\le a_m$.

Step 2: convergence. Let $\varepsilon>0$. By clause 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is an element of $A$ that is greater than $s-\varepsilon$; every element of $A$ is of the form $a_N$ for some $N\in\mathbb{N}$, so $s-\varepsilon<a_N$ for some $N\in\mathbb{N}$. Let $n\in\mathbb{N}$ with $n\ge N$. By Step 1, $a_N\le a_n$, and $a_n\le s$ because $s$ is an upper bound for $A$. Hence

$$s-\varepsilon<a_N\le a_n\le s<s+\varepsilon ,$$

so $-\varepsilon<a_n-s\le 0$ and therefore $|a_n-s|<\varepsilon$. Since $\varepsilon>0$ was arbitrary, $(a_n)$ converges to $s$.

2. (Nonincreasing case.) Assume $a_{n+1}\le a_n$ for every $n\in\mathbb{N}$ and that $A$ is bounded below. Since $A$ is nonempty and bounded below, $t=\inf A$ exists by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below.

Step 1: $a_m\le a_n$ whenever $n\le m$. Fix $n\in\mathbb{N}$ and let $T'=\{m\in\mathbb{N}: m<n\ \text{ or }\ a_m\le a_n\}$. As before $1\in T'$: either $1<n$, or $n=1$ and $a_1\le a_1$. Suppose $m\in T'$. If $m<n$, then $m+1\le n$, so either $m+1<n$ or $m+1=n$, and in the latter case $a_{m+1}\le a_n$; in both cases $m+1\in T'$. If $n\le m$, then $a_m\le a_n$ because $m\in T'$, and $a_{m+1}\le a_m$ by hypothesis, so $a_{m+1}\le a_n$ and $m+1\in T'$. By induction $T'=\mathbb{N}$, which gives the claim.

Step 2: convergence. Let $\varepsilon>0$. By clause 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $N\in\mathbb{N}$ with $a_N<t+\varepsilon$. Let $n\ge N$. By Step 1, $a_n\le a_N$, and $t\le a_n$ because $t$ is a lower bound for $A$. Hence

$$t-\varepsilon<t\le a_n\le a_N<t+\varepsilon ,$$

so $|a_n-t|<\varepsilon$. Since $\varepsilon>0$ was arbitrary, $(a_n)$ converges to $t$.