Monotone Sequences converge to their Extremum

Last updated Nov 1, 2022

$(x_{n}) \subset \mathbb{R}$ be a Monotone Sequence. Let $t \in \bar{\mathbb{R}}$ be it’s Extremum corresponding to the direction of the monotone sequence ( if Non-Increasing Function then Infimum; if Non-Decreasing Function then Supremum). Then $x_{n} \to t$.

First suppose that $(x_{n})$ is Non-Decreasing Function. Observe that $t = \sup{x_{n}}$ exists in $\bar{\mathbb{R}}$ since either

1. $(x_{n})$ is bounded from above in $\mathbb{R}$: By Completeness of the Real Numbers, we know $\sup{x_{n}} \in \mathbb{R}$.
2. $(x_{n})$ is not bounded from above in $\mathbb{R}$: $\infty$ is always an Upper Bound (by definition), so $\sup{x_{n}}$ exists. Since no other element of $\bar{\mathbb{R}}$ is an Upper Bound (we’re not bounded in $\mathbb{R}$ and $- \infty$ is less than every element in $\mathbb{R}$), $\infty$ is the only upper bound and thus the Supremum.

First we prove the statement for when $t \in \mathbb{R}$. Now let $\epsilon > 0$. Then because Finite Extremums get arbitrarily close, we know there exists $N \in \mathbb{N}$ so that $x_{N} > t - \epsilon$. Since $(x_{n})$ is non-decreasing and $t = \sup{x_{n}}$, we know for all $n \geq N$: $$t \geq x_{n} \geq x_{N} > t - \epsilon$$ Thus, for all $n \geq N$: $$|t - x_{n}| = t - x_{n} < \epsilon$$ We’ve satisfied the definition of Sequence Convergence, so we have that $(x_{n}) \to t$.

Now suppose $t = \infty$. Then we know $(x_{n})$ has no Upper Bound in $\mathbb{R}$. In other words, $\forall M \in \mathbb{R}$, we know there exists $N \in \mathbb{N}$ so that $x_{N} > M$. Since $(x_{n})$ is Non-Decreasing Function, we have that for all $n \geq N$ $$x_{n} \geq x_{N} > M$$ This satisfies the definition of Sequence Convergence to Infinity, so $(x_{n}) \to \infty$ .

Now consider when $(x_{n})$ is Non-Increasing Function. Then $\forall n > m \in \mathbb{N}$, we have that $$x_{n} \leq x_{m}$$ Thus, $$-x_{n} \geq -x_{m}$$ and $(-x_{n})$ is a Non-Decreasing Function sequence. By our proof above, we know

$$(-x_{n}) \to \sup{-x_{n}} = -\inf{x_{n}}$$ where the last equality follows because Supremum of Negative is Negative of Infimum. Since $x \mapsto -x$ is Continuous Function, we know $(x_{n}) \to \inf{x_{n}}$. $\blacksquare$

• Real Numbers
• Extended Real Numbers
• Sequence Convergence