$\sup A = \inf B$ implies $\forall\varepsilon>0.\exists a\in A, b\in B. b-a<\varepsilon$

Question

Let $A, B$ two sets such that $\sup A = \inf B$. Is it right that:
$$ \forall \varepsilon > 0. \exists a\in A, b\in B. b-a<\varepsilon \quad ? $$

The question doesn't mention the sets are densed, but that was probably the intention.

I think the claim is true.
for every $\varepsilon > 0$ we are able to choose $a\in A$ such that $\sup A - a < {\varepsilon \over 3}$ and $b\in B$ such that $b - \inf B < {\varepsilon \over 3}$.

And so, the distance is ${2 \over 3\varepsilon} < \varepsilon$.

We can choose $a, b$ as mentioned because those sets are densed.

Is that right? (Or am I being silly here? because the proof is almost trivial).

Answer 1

You're correct, but being a bit silly. The thing you are being silly about is the "denseness" thing. This is unnecessary; definition of the supremum and infemum imply that for any positive $\epsilon$, there is an element at most $\epsilon$ away from the supremum (or infemum) in the set. (This element may, of course, be equal to the supremum or infemum itself.) In particular this is true for $\frac{\epsilon}{3} > 0$.

Also note Hagen von Eitzen's comment: you must assume the supremum and infemum are finite real numbers for this to work.

Answer 2

Yes, that's right. You can always make that choice, by definition of $\sup A$ (if that's not a maximum then you can choose elements at distance less that $\varepsilon$, else $supA-\varepsilon$ would be a smaller $\sup A$). FALSE( Denseness is needed if you have to choose an element with distance exactly $\varepsilon$ )