[Math] Show that a sequence is bounded if and only if there exists a K $\in\mathbb{R}$ such that $|\ a_n\ | \leq K$ $\forall n\in \mathbb{N}$.

real-analysissequences-and-series

Show that a sequence is bounded if and only if there exists a K $\in\mathbb{R}$ such that $|\ a_n\ | \leq K$ $\forall n\in \mathbb{N}$.

Is it correct/enough for me to show the following:

  1. Let $K \in \mathbb{R}$ and assume $| a_n | \leq K$ for all n $\in \mathbb{N}$. Then from basic properties of absolute inequalities, we know that $-K \leq a_n \leq K$, and hence the sequence is bounded from above and below and hence the sequence is said to be bounded.

  2. Assume the sequence is bounded (from above and below), which tells us
    $\exists \ m,M$ $\in \mathbb{R}$ such that $m \leq a_n \leq M$ for all $n \in \mathbb{N}$. Now, since both $m,M$ are unknown variables, if we choose a value $K \in \mathbb{R}$ such that $m=-K \ \text{and} \ M=K$ , then we can rewrite this as $|a_n| \leq K$. Thus proving our statement.

EDIT: This is the given definition of bounded:

A sequence ($a_n$)

a) is bounded from above if there exists a M $\in \mathbb{R}$ such that $a_n \leq M$ for all n

b) is bounded from below if there exist a m $\in \mathbb{R}$ such that $m\leq a_n$ for all n.

c) is bounded if both the above holds

Best Answer

1) is correct and for 2) let $$K=\max(|m|,|M|)$$ then

$$-K\le m\le |a_n|\le M\le K$$

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