Let there be $A,B\in \mathbb{R}$ non-empty, let assume that $\forall a\in A \exists b \in B$ such that $a\leq b$ and $\forall b\in B \exists a \in A$ such that $b\leq a$.

Show that $A$ is bounded from above if and only if $B$ is bounded from above and than $\sup(A)=\sup(B)$.

We know that $a\leq b$ for all $a\in A$ therefore $b$ is an upper bound of $A$ and the same goes to $b\leq a$ so a is an upper bound of $B$.

But $a\leq b$ and $b\leq a$ so $a\leq b\leq a. $ How should I proceed to show that $\sup(A)=\sup(B)$?

  • $\begingroup$ @Regret is I know because of that the sets can be $A\neq B$ $\endgroup$ – gbox Mar 30 '15 at 12:24
  • $\begingroup$ Your first step isn't correct. We know that $\forall a \in A:\exists b \in B: a\leq b$. You can't invert $\forall a$ and $\exists b$. $\endgroup$ – Kitegi Mar 30 '15 at 12:55

You argue something like this:

If $A$ is bounded from above then $\exists M$ such that $\forall a \in A$, $a \leq M$. Since for all $b \in B$ there exists an $a \in A$ such that $b \leq a \leq M$ then we can conclude that $B$ is also bounded from above. We would also need to write down the reverse argument to show that $B$ bounded from above implies $A$ bounded from above.

Now if $sup(A)$ is the supremum of $A$ then it means it is the smallest upper bound for $A$. Now since we have for all $a \in A$, $\exists b \in B$ such that $a \leq b \leq sup(B)$, we see that $sup(B)$ is also an upper bound for $A$ and so $sup(A) \leq sup(B)$, (remember $sup(A)$ is the smallest lower bound). We write down the completely symmetric argument to get $sup(B) \leq sup(A)$. We therefore can conclude that $sup(A)=sup(B)$.

  • $\begingroup$ If the OP understands the basic definitions in this stuff of upper/lower bounds, supremums and etc. , I think this is the simplest answer possible. +1 $\endgroup$ – Timbuc Mar 30 '15 at 12:28

You have to remember that with for any $a \in A$, $a \leq \sup{A}$. Also, for any $b \in B$, $b \leq \sup{B}$.

Finally, remember that if there is some $M$ such that for all $a \in A$, $a \leq M$, then we get $a \leq \sup{A} \leq M$ since supremum is the least upper bound (i.e., if $M$ is an upper bound of $A$, then the sup would be smaller than this $M$).

Can you use what I said above to prove $\sup{A} = \sup{B}$? In particular, if for every $a \in A$, there is some $b \in B$ such that $a \leq b$, how does $\sup{B}$ fit into this equality? What does that make $\sup{B}$ with respect to the set $A$? What does that imply about the relationship of $\sup{A}$ and $\sup{B}$? The answers to all of these questions can be figured out from the two things I had you recall in the beginning of this answer.

You can ask the exact same questions, except swapping $\sup{A}$ and $\sup{B}$, with the other statement about all $b \in B$.


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