Metric space bounded property

Question

According to classical definition In $R $ we know that a set $S$is bounded if there exist $k$$\in $$R $ such that $|a|$$<$$k $ for all $a$$\in $$S$. But according to what I have read in metric space so far now a set is bounded if $diam (S) $ is finite. Is there any way to prove they are equivalent??

I have tried to prove this by taking $S$ is bounded. And then proving that if we map a function $d$ from $S$ to $d(S)$($d$ follows standard conditions of a function being metric). And then prove that $d (S)$ is bounded. We can prove it. But I am in the dark through this approach. So any kind of help is appreciated.

Answer 1

I think you mean $\text{diam}(S)$ rather than $\text{dim}(S)$, these mean very different things. The usual definition of the diameter of a set in a metric space is $$ \text{diam}(S)=\sup_{a,b\in S}d(a,b). $$ Here the metric under consideration is $d(a,b)=|a-b|$, the Euclidean metric on $\mathbb R$.

To show that the two notions of boundedness are the same, we will show that (1) if $\text{diam}(S)<\infty$, then $S\subseteq [-M,M]$ for some $M\geq 0$, and (2) if $S\subseteq[-M,M]$, then $\text{diam}(S)<\infty$.

(1) Take any point $x\in S$. By definition of the diameter, $S\subseteq [x-\text{diam}(S),x+\text{diam}(S)]$. This, in turn, is contained in the interval $[-M,M]$ if we set $M=|x|+\text{diam}(S)$.

(2) If $S\subseteq[-M,M]$, then $\text{diam}(S)\leq \text{diam}([-M,M])=2M<\infty$.

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Edit in response to further question: As stated, the question only makes sense for $\mathbb R$, because it is not clear what one means by the absolute value of an element in an arbitrary metric space. However, there is a straightforward generalization if one considers a metric space induced by a norm. This consists of a vector space $S$ equipped with a norm function mapping $x\in S$ to a non-negative real number $\|x\|\geq 0$ such that $\|ax\|=|a|\|x\|$ for all real numbers $a$ and such that $\|x+y\|\leq \|x\|+\|y\|$ and $\|x\|=0$ implies $x=0$. Then the metric is defined to be $d(x,y)=\|x-y\|$. For any metric space induced by a norm in this manner, the same proof as above shows that the diameter of a set is finite if and only if the norm is bounded on that set.

Answer 2

This is really an extended comment to the other answer, rather than an answer.

One possible way to define $|\cdot|$ in a general metric space is to pass from metric spaces to pointed metric spaces. In topology, a pointed space is a topological space $X$, along with a point $x \in X$ which is singled out as the 'base point' of the space. Frequently in topology it is important to use spaces with a chosen base point rather than just spaces because we want to associate some further data to the space, such as the fundamental group of the space (although one can prove that the fundamental group is the same regardless of the choice).

Now in your case, if you choose a base point $x$ for the metric space, it will make sense to talk about $| \cdot |$ as just $d(x, \cdot)$. This is a natural generalization of what happens in $\mathbb{R}^n$, where the base point is just $0$.

Now in this setting, if $S \subseteq X$ is bounded, then by your definition $\sup_{a,b \in S} d(a,b)$ is finite. I think it's clear that this implies $|a|$ is bounded. On the other hand, if $|a|$ is bounded, then for any pair $a,b$ we can write $d(a,b) \leq |a| + |b|$ by triangle inequality. (Note that is still doesn't make sense to write $|a+b|$ as in a general metric space we have no notion of adding or subtracting points, unless perhaps we're in a vector space like in the other answer.) Since $|\cdot|$ is a bounded function $X \to \mathbb{R}$, it follows that the diameter is bounded by twice whatever the sup of $|\cdot|$ is.