Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I see quite a lot of different definitions of a bounded space. For instance, from nLab:

Let $E$ be a metric space. A subset $B⊆E$ is bounded if there is some real number $r$ such that $d(x,y)<r$ for all $x,y∈B$.

From Wiki:

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 understand it correctly, the first definition requires that the origin of the "open disc" must be inside the subset, whereas the second definition does not have this restriction. Are these definitions somehow the same, or are they different? If not which one is correct?

The motivation for this question is because I want to understand what the Heine-Borel Theorem means.

share|improve this question

4 Answers 4

up vote 3 down vote accepted

They are the same. Say you have any $x$ with $M \subset B_r(x)$, where $B_r(x) = \{y \,:\, d(x,y) < r\}$. Then pick an arbitrary $z \in M$. Due to the triangle inequality, you have $$ M \subset B_{r + d(x,z)}(z) \text{.} $$

share|improve this answer

Assuming the subset is nonempty, the two definitions are equivalent. Each implies the other.

share|improve this answer
3  
But the $r$ in one definition may need to be $2r$ for the other. And, as noted, one definition is wrong for $M=\varnothing$. –  GEdgar Mar 28 at 18:13

They're equivalent. That's wonderful because it means that you get to use the most convenient definition for whatever you're trying to do. As an exercise, you should prove that each implies the other.

share|improve this answer

The two definitions are equivalent, let me give a quick proof:

Let $B\subseteq E$ be a subspace of a metric space $(E,d)$.

$\Rightarrow$: Assume there is an $r>0$ such that $d(x,y)<r$ for all $x,y\in B$. Fix any $p$ in $B$ then $d(p,x)<r$ for all $x\in B$.

$\Leftarrow:$ Assume there is a $p\in E$ and $r>0$ such that $d(p,x)<r$ for all $x\in B$ then for all $x,y\in B$ we have $$d(x,y) \le d(p,x) + d(p,y) < r+r=2r.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.