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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?

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

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  • $\begingroup$ I see, can you please give a hint on how to prove this problem? $\endgroup$ Jan 7 '17 at 21:22
  • $\begingroup$ 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 $B$. $\endgroup$
    – Yorch
    Jan 7 '17 at 21:25
  • $\begingroup$ 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? $\endgroup$ Jan 7 '17 at 21:33
  • $\begingroup$ $lub(A)$ 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 $b\in B$. $\endgroup$
    – Yorch
    Jan 7 '17 at 21:36
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If $A$ is not bounded above, you cannot have $a\le b$ for all $a\in A$ ($b$ fixed).

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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 $b\in B$, $a\leq b$.

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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)

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  • $\begingroup$ 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. $\endgroup$ Jan 7 '17 at 21:55
  • $\begingroup$ @user3000482 No that doesn't work. It proves too much: some a is larger than the infimum of B, not necessarily all a. $\endgroup$ Jan 7 '17 at 22:06
  • $\begingroup$ 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)? $\endgroup$ Jan 7 '17 at 22:12
  • $\begingroup$ @user3000482 But remember we were in proof by contradiction land. You switched back to the original domain of discourse. $\endgroup$ Jan 7 '17 at 22:15

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