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Let $d$ be the usual metric on $\mathbb{R}$. A subset $A$ of $(\mathbb{R},d)$ is bounded if and only if it has both a lower bound and an upper bound.

Many times I have seen this paragraph as a definition but I would like to see a proof of this please.

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closed as not a real question by Andres Caicedo, Hagen von Eitzen, Ittay Weiss, Davide Giraudo, Thomas Jan 6 '13 at 15:35

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What do you mean by you want to see a proof of a definition? –  Calvin Lin Jan 4 '13 at 2:19
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What do you propose to use as the definition of bounded? –  Gerry Myerson Jan 4 '13 at 2:22
    
We'd be happy to prove a theorem a set is bounded if and only if it has an upper and a lower bound, provided you give us a nice definition of "bounded". –  Hagen von Eitzen Jan 6 '13 at 11:36

3 Answers 3

I will prove that there is some $r>0$ such that $d(a,b)<r$ for all $a,b\in A$ if and only if $A$ has an upper bound and a lower bound.

Suppose that $A$ is bounded, then there is some $n$ such that $d(a,b)<n$ for all $a,b\in A$. Then $A\subseteq[a-n,a+n]$ for some $a\in A$, and therefore has lower and upper bounds.

In the other direction, if $A$ has a lower bound $x$, and an upper bound $y$ then $A\subseteq[x,y]$ and therefore $d(a,b)<y-x$ by the triangle inequality. Therefore $A$ is bounded.

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Good, thanks for you answer. –  Fernando Valle Jan 4 '13 at 2:40

Let the lower bound be $x$ and the upper bound be $y$. Then the ball centred at $0$ with radius $\max(|x|,|y|)+1$ contains $A$.

Conversely, if there is a ball centred at $z$ with radius $r$ that contains $A$, then a lower bound is $z-r$ and an upper bound is $z+r$.

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The "usual" (I assume) definition of "bounded" is that a subset $A\subseteq \mathbb{R}$ is bounded if and only if there exists $M\in \mathbb{R}$ such that $\left|a\right|\leq M$ for all $a\in A$ (where $\left|a\right|$ denotes the absolute value of $a$). In particular, $M$ is an upper bound of $A$. Also, $-\left|a\right|\geq -M$ for all $a\in M$ which implies that $-M$ is a lower bound of $A$. Therefore, $A$ has both an upper bound and a lower bound.

I hope this helps!

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