Why is the $\operatorname{diam}{(\mathbb Q)}$ infinite?

Question

I was trying to find a counterexample to show that the $\operatorname{diam}{(A)}$ and the $\operatorname{diam}{(Int(A))} $ may not be the same, where $A$ is the subset of the metric space $X$.

I chose the metric space $X$ = $\mathbb R$ , and chose $A$ = $\mathbb Q$.

So, since $Int(\mathbb Q)$ is empty , i concluded that the $\operatorname{diam}{(Int(\mathbb Q))}$ = $0$.

Then, i checked up online, to find that the $\operatorname{diam}{(\mathbb Q)}$ is infinity.

So, i think that my counterexample would probably work.

I do understand intuitively, why the $\operatorname{diam}{(\mathbb Q)}$ is infinity,

What i wanted to know is that if we were to prove this then,

we would probably have to show that $a-b$ > $M$ where $a,b \in \mathbb Q$ and $M$ is an

arbitrarily large positive real number.

Am i correct ?

It would be nice if someone could shed some light on this !

Answer 1

It is probably better to make an example that doesn't involve infinity or the empty set. You could take $$ A=(\mathbb Q\cap[0,1])\cup[0,1/2]. $$ Then $$ \mbox{diam}(A)=1,\ \ \mbox{diam}(\mbox{int}(A))=1/2. $$ And of course, you can take $d$ to be any other number in $(0,1)$.

As for your question, think of the diameter of the integers (which are part of the rationals).

Answer 2

The diameter of $\mathbb Q$ is $\sup\{|b-a|:a,b\in\mathbb Q\}$. Can you bound that above by anything?