If $\inf(A) < \sup (B) $ then there is some $a \in A$ and $b \in B$ such that $a<b$

Question

Proposition: Let A be a subset of R which is bounded below. Let B be a subset of R which is bounded above. If $\inf(A) < \sup (B) $ then there is some $a \in A$ and $b \in B$ such that $a<b$.

Proof:

Let $\inf(A) = I$ and $\sup(B) = S$

So $I<S$

By definition, $\forall a \in A, I\leq a$ and $\forall b \in B, b\leq S$

Case 1: If I $\in A$ and S $\in B$, we are done.

Case 2: If I $\notin A$ and S $\notin B$, then assume BWOC $\nexists a \in A $ and $b\in B$ such that $a<b$

So $\forall a \in A$ and $\forall b \in B, b \leq a$

So $\forall a \in A$, $a$ is an upper bound for B.

But since S is the least upper bound for B, $b\leq S<a$

$S<a$

So $S$ is also a lower bound for $A.$

But since $I$ is the greatest lower bound for $A$, $S<I$, which is a contradiction.

Is my proof valid? If not, what am I missing? Would I have to show that this holds for other cases as well?

Answer 1

Your proof works fine; note that what you've done in case $2$ is actually sufficient to prove the full statement, though. As you say, assume that there do not exist elements $a\in A$ and $b\in B$ such that $a<b$. Then for any $a\in A$, we have that $a\geq b$ for any $b\in B$. In particular, $a$ is an upper bound for $B$ so that $\sup B\leq a$. Since $a\leq\sup A$ by definition, we have $\sup B\leq \inf A$, and this is a contradiction.

In fact, you don't need to use contradiction; the statement is not too hard to prove directly. Define $\varepsilon=\sup B - \inf A$. By definition of infimum and supremum, there exists an $a\in[\inf A,\inf A+\varepsilon/2)$ and similarly there exists a $b\in(\sup B-\varepsilon/2,\sup B]$. By construction, $a<b$ so the statement is proven.

Answer 2

Your proof works, but there is a little flaw: you're not considering the case $I \in A, S \not\in B$ or viceversa; by the way, dividing in cases is useless cause what you do in the last case can be modified to work in every case. You can simply change $S<a, S<I$ into $S \le a, S \le I$ and you'll obtain a contradiction anyway.

Answer 3

I think it can be proved without examining different cases:

By definition of $\;\inf A$, for any $\varepsilon >0$, there exists $a\in A$ such that $$I\le a <I+\varepsilon.$$

Similarly, for any $\varepsilon >0$, there exists $b\in B$ such that $$S-\varepsilon < b \le S.$$ Now, if $\;I+\varepsilon <S-\varepsilon$, you're done. Do you see how this can be realised?