Metric Spaces don't understand this question in Robert Magnus' book

Question

I am going through Metric spaces by Robert Magnus and in Chaper 1.1 he states the following: "Let $(X, d)$ be a metric space. A subset $A$ of $X$ is said to be bounded if there exists a point $a ∈ X$ and $R > 0$, such that for all $x ∈ A$ we have $d(a, x) < R$."

Then he says: "Fix a point $b ∈ X$. Show that a set $A$ is bounded if and only if there exists $R > 0$, such that for all $x ∈ A$ we have $d(b, x) < R$. The point here is that the same point $b$ can be used to test sets for boundedness, independent of the set in question."

In this question even after his little explanation at the end I still don't understand what he is asking me to do. And I don't really understand how to turn the variable point "$a$" into a fixed point "$b$". I also don't really understand how to make a set A with constraints to a set A without the constraints of being in X. Basically the whole question is a big "?" for me at the moment...If somebody could give me any hints on how to solve this little question that would be very appreciated :)

Answer 1

Let's call such a point $a$ as described in the definiton of "$A$ is bounded" a 'witness' for $A$'s boundedness. If a set is bounded, we know there must be at least one witness.

If you have a set $A$ that you want to check for boundedness, you have to 'guess' a suitable point $a$ and then check the set of distances

$$D(a)=\{d(a,x)|x \in A\}.$$

If that set is bounded (in the classical sense on $\mathbb R$), you are done, as your guessed 'a' is indeed a witness, so $A$ is bounded by definition.

But what happens if that set is not bounded? Then, in theory, you would have to guess another point $a'$, look at $D(a')$ and if it's bounded, you are done ($a'$ is a witness) and if not, you have to guess again and again (which is impossible to do completely for metric spaces with infinitely many points).

What you are asked to prove is that this scenario cannot happen. If you want to check if a set $A$ is bounded, you can take any point $b \in X$ and calculate the distance set $D(b)$. If it is bounded, $b$ is a witness and $A$ is obviously bounded. But if $D(b)$ is not bounded, then (you are asked to prove) $D(b')$ is unbounded for any $b' \in X$, that means there are no witnesses at all in $X$, so $A$ cannot be bounded.

So if a set $A$ is bounded, any point in $X$ works as a witness for that. That makes checking for boundedness much easier, you only have to look at one point $p$ in $X$ and calculate $D(p)$, instead of potentially needing the check all of them in case $D(p)$ is unbounded.