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I'm looking at the definition of the bounded set on Wikipedia, which says:

A subset S of a metric space $(M, d)$ is bounded if it is contained in a ball of finite radius, i.e. if there exists $x \in M$ and $r > 0$ such that for all $s \in S$, we have $d(x, s) < r$.

If I let $M = \mathbb{R}$ and $S$ be the set of all nonnegative real numbers, does that mean it's not bounded by the definition? I'm struggling to construct a ball to contain $S$.

If so, that's a bit counter-intuitive to me, since $S$ is bounded from below (by $0$). Can someone please verify whether or not I'm right?

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2 Answers 2

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Let $a\in S$. Let $r > 0$.

Then $B_r(a)$ is the "ball of radius $r$ around $a$", and in this case it corresponds to the interval $(a-r, a+r)$.

But clearly $a+r+1 \in S$, while $a+r+1 \not\in B_r(a)$. Because $a$ and $r$ were arbitrary, this shows that $S$ isn't contained in any ball, so it's not bounded.

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As you say, $S = [0,+\infty)$ is bounded from below. This is weaker than being bounded, and the easiest example to see this is actually $S$. If it were bounded, we could take an interval $(-M,M)$ with $M > 0$ such that $S \subseteq (-M,M)$. This is false, since $M+1$ is in $S$ but not in $(-M,M)$. Therefore, $S$ is not bounded.

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  • $\begingroup$ I meant that to be a parenthesis, I've edited it accordingly. Thanks! $\endgroup$
    – qualcuno
    Nov 5, 2017 at 6:07

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