# On the proof that if $a \leq b$ for all $a \in A$ and $b \in B$ then $\sup(A)\leq\inf(B)$

Suppose $A$ and $B$ are subsets of $\mathbb{R}$, both nonempty, with the speical property that $a \leq b$ for all $a \in A$ and for all $b \in B$. Prove: sup$(A)$ $\leq$ inf$(B)$.

I have an objection to this problem. What if $A$ is a monotone increasing sequence that is not bounded above and $B$ is also monotone increasing sequence that is not bounded above but bounded below? Then sup$(A)$ $= \infty$, inf$(B)=M$ for some $M \in \mathbb{R}$. Hence, the inequality won't make sense. Do I have to assume that these subsets are bounded?

• supA <=b for all b. Thus supA is <=infB. Jan 7 '17 at 21:54

if $A$ is not bounded above then the sets $A$ and $B$ cannot fulfill the required conditions, because if we pick $b\in b$ then this $b$ is not an upper bound for $A$, and hence there is $a\in A$ with $b<a$.

Another way to verify that your counterexample does not work is to prove the theorem.

• I see, can you please give a hint on how to prove this problem? Jan 7 '17 at 21:22
• notice that every element of $b$ is an upper bound for $B$. This implies that the least upper bound of $A$ is a lower bound for $B$, and therefore the least upper bound of $A$ must be smaller than the greatest lower bound of $B$. Jan 7 '17 at 21:25
• How is it implied that lub($A$) is the lower bound for $B$? I am sorry, it makes sense but could you show this with mathematical rigours? Jan 7 '17 at 21:33
• $lub(A)$ is a lower bound for $B$. Because every element of $b$ is an upper bound of $A$. Since $lub(A)$ is the smallest upper bound of $A$ we conclude that $lub(A) \leq b$ for all $b\in B$. Jan 7 '17 at 21:36

If $A$ is not bounded above, you cannot have $a\le b$ for all $a\in A$ ($b$ fixed).

Suppose $A$ and $B$ are subsets of $\mathbb{R}$, both nonempty, with the speical property that $a \leq b$ for all $a \in A$ and for all $b \in B$. Prove: sup$(A)$ $\leq$ inf$(B)$.

Steps for the proof

• Since $B\neq\emptyset$, let $b$ be some element of $B$, show that $$\sup(A)\leq b\tag{1}.$$
• Show that (1) is true for any $b\in B$. It follows that $\sup(A)$ is a lower bound for $B$.
• Finish the proof.

I have an objection to this problem. What if $A$ is a monotone increasing sequence that is not bounded above and $B$ is also monotone increasing sequence that is not bounded above but bounded below? Then sup$(A)$ $= \infty$, inf$(B)=M$ for some $M \in \mathbb{R}$. Hence, the inequality won't make sense. Do I have to assume that these subsets are bounded?

Exercise: Show that by the assumptions of the problem, $A$ must be bounded above and $B$ must be bounded from below.

HINT: Since for any $a\in A$ and any $b\in B$, $a\leq b$, it is in particular true that for any $a\in A$ and some $b\in B$, $a\leq b$.

If sup A > inf B then a>inf B for some a, and thus a>b for some b (for if it wasn't then inf B would not be the infimum of B)

• Is this valid? If sup(A)>inf(B), then we can have $\epsilon =$sup(A)-inf(B) $> 0$. Now, sup(A)-$\epsilon$ cannot be an upper bound for $A$ because sup(A) is the least upper bound. Hence, there is some $a \in A$ such that sup(A)-$\epsilon \leq a \leq$ sup(A), which is equivalent to inf(B)$\leq a \leq$sup(B). But a $\geq$ inf(B) for all a. Jan 7 '17 at 21:55
• @user3000482 No that doesn't work. It proves too much: some a is larger than the infimum of B, not necessarily all a. Jan 7 '17 at 22:06
• Isn't it true that for all $a \in A$, $a \leq$inf(B)? Because every $a$ is a lower bound for $B$ and inf(B) is greatest lower bound? Hence, $a \leq$ inf(B)? Jan 7 '17 at 22:12
• @user3000482 But remember we were in proof by contradiction land. You switched back to the original domain of discourse. Jan 7 '17 at 22:15