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## Proof: (related to "A Criterion for Subsets of Real Numbers to be Bounded")

### $”\Rightarrow”$

• Let $D$ be bounded.
• By definition, $D$ is bounded below, i.e. exists a real number $L$ with $x\ge L$ for all $x\in D.$
• Also by definition, $D$ is bounded above, i.e. exists a real number $U$ with $x\le U$ for all $x\in D.$
• Without loss of generality, we can assume that $U > 0.$
• Set $M:=\max(|U|,|L|)$ (maximum of both absolute values.)
• Since $M > 0,$ there is a positive real number $M$ with $|x|\le M$ for all $x\in D.$

### $”\Leftarrow”$

• Let $M > 0$ be a positive real number $M$ with $|x|\le M$ for all $x\in D.$
• Since $-M < x$ for all $x\in D,$ the set $D$ is bounded below.
• Since $M > x$ for all $x\in D,$ the set $D$ is bounded above.
• By definition, $D$ is a bounded set.
q.e.d

| | | | created: 2020-03-17 18:44:04 | modified: 2020-03-17 18:44:19 | by: bookofproofs