Are there any subsets $A, B$ of $\mathbb R$ such that $\sup(B)>\sup(A)$ but none of the elements of $B$ are upperbounds of $A$?

Question

If not, then how would you prove it? I've thought of multiple examples that do not support this but cannot think of how to go about proving it.

Answer 1

Suppose $b:=\sup B>\sup A=:a $. Then for any $\varepsilon>0$ we have $B\cap(b-\varepsilon,b]\neq\varnothing$. Since $b>a$ we can choose $\varepsilon$ such that $b-\varepsilon>a$, and thus $B$ has an element greather than $a$, and this is an upper bound for $A$.

Answer 2

Now I need the fact that $\sup(A),\sup(B)\in \mathbb{R}$ by completeness of $\mathbb{R}$.

Then since $\sup{A}<\sup{B}$, so $\exists b^*\in B$ such that $b^*>\sup{A}$ (by definition of sup).

Hence, $b^*$ is the desired upper bound.

Answer 3

The assumption: $\forall b\in B \exists a\in A : a\gt b$.

But, $\forall a\in A, a\le supA$, by definition.

So $\forall b\in B, supA\ge b$.

Thus $supA\ge supB$...

Answer 4

When all else fails, try thinking about the definitions. Here's how you do that.

$\sup B\gt\sup A$

$\sup B$ is the LEAST UPPER BOUND of $B.$

Sinse $\sup B$ is the LEAST upper bound of $B,$ nothing smaller than $\sup B$ can be an upper bound of $B.$

Since $\sup A$ is smaller than $\sup B,$ that means $\sup A$ is not an upper bound of $B.$

Since $\sup A$ is not an upper bound of $B,$ that means some element of $B$ is bigger than $\sup A.$

Since $\sup A$ is an upper bound of $A,$ that means some element of $B$ is bigger than an upper bound of $A.$

Since anything bigger than an upper bound is also an upper bound, that means some element of $B$ is an upper bound of $A.$