A is bounded iff $\forall x \in M: \sup_{a \in A}d(x,a)< \infty$

Question

I want to proof the following, given a metric space $(M,d)$ and a subet $A$ of M.

A is bounded iff $\forall x \in M: \sup_{a \in A}d(x,a)< \infty$

1). FLTR. Let $x$ be an arbitrary element of $M$. We're going to show that for any $a \in A$, that $d(x,a) < \infty$ holds, thus it also holds for: $\sup_{a \in A} d(x,a)$. However, it is per definition that for any two element of $M$ have a distance from eachother that is finite. So we're finished

2). FRTL. We want to show that $A$ is bounded, that is, for some $y \in M$ there is an $r>0$ such that $A \subset B_r(y)$. We choose $x \in A$, $x \in M$ so thus, $C = \sup_{a \in A}d(x,a)<\infty$ per the antecedent. But then, $A \subset B_C(x)$ for some $x\in M$ and some $C >0$.

I feel as if especially proof 1). is not quite okay. I believe that the conclusion that it holds for all $a \in A$ to that it also holds for the supremum, is not correct. Could someone help me out?

Thanks,

K. Kamal

Answer 1

1) You can't deduce from the fact that each number is $<\infty$ that $\sup$ also has the same property.

Suppose that $A$ is bounded. Then $A\subset B_r(y)$, for some $y\in M$ and some $r>0$. Fix $x\in M$. Then, for each $a\in A$,$$d(x,a)\leqslant d(x,y)+d(y,a)<d(x,y)+r$$and therefore$$\sup\{d(x,a)\,|\,a\in A\}\leqslant d(x,y)+r.$$

2) There is no need to choose $x\in A$. On the other hand, it doesn't have to be true that $A\subset B_C(x)$. Take some $r>C$ and then, yes, $A\subset B_r(x)$.

Answer 2

If $A$ is bounded, then there exists a ball ${B(p,r)}$ that contains $A$. So for any $x\in M$, we have $$ \sup_{a\in A} d(x,a) \leqslant \sup_{a\in A} (d(x,p) + d(p,a)) \leqslant d(x,p) + r<\infty. $$

Conversely, suppose $\rho_x:=\sup_{a\in A}d(x,a)<\infty$ for all $x\in M$. Let $a\in A$, then $d(a,a')<\rho_a$ for all $a'\in A$. It follows that $A\subset B(a,\rho_a)$, and hence $A$ is bounded.