Given two bounded, nonempty sets $A$ and $B$, such that $\sup(A) = \inf(B)$.
$(a)$ Show that $A \cap B$ contains at most $1$ element.
Suppose that two distinct $a, b \in A \cap B $, we then have:
$$ a \leq \sup(A) = \inf(B)$$
since $a \in A$, and also since $b \in B$ we can say that that:
$$ a \leq \sup(A) = \inf(B) \leq b \rightarrow a \leq b$$
We now make the same statement, but reverse the roles of $a$ and $b$, since $b \in A$, and also since $a \in B$
$$ b \leq \sup(A) = \inf(B) \leq a \rightarrow b \leq a$$
We conclude that $a=b$. $\square$
I think this proof is fine so far.
(b) Show that:
$$ \forall \epsilon >0: \space{ } \exists a \in A, b \in B: \space |a-b|< \epsilon $$
I think the point is to use the first exercise,
1 element so suppose we have $c \in A \cap B$ then also $c \in A,B$ , we then have that $|c-c|=0<\epsilon$
No elements For the second case we would need to prove that whenever $A \cap B = \emptyset$, we can still find such an $a$ and $b$. Maybe we could somehow use the fact about the supremum and infimum, but I am not sure how to use this. We know that $a \leq \sup(A)=\inf(B) \leq b$, this just tells us how they are ordered.
Best Answer
I think it's unnecessarily complicated to use the first question. It could be useful to use the following property of $c=\sup(A)$. For every $\varepsilon>0$ there exists $a\in A$ such that $$c-\varepsilon<a$$ (otherwise $c$ wouldn't be the lowest upper bound!). Similarly if $d=\inf(B)$ then for every $\varepsilon>0$ there exists $b\in B$ such that $$b<d+\varepsilon$$ Can you finish it from there?