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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 Andrés E. Caicedo, Hagen von Eitzen, Ittay Weiss, Davide Giraudo, Thomas Jan 6 '13 at 15:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What do you mean by you want to see a proof of a definition? – Calvin Lin Jan 4 '13 at 2:19
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

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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