Supremum Definition

Question

I am using this and am looking at problem 2.

When looking at the solution for 2, it states that since $B - \epsilon < $sup $Y$ (and B is defined as sup Y), then there exists an element y in Y such that sup $Y - \epsilon < y$... how does this follow? and how do you know such an element exists?

Answer 1

The supremum of $Y$ is the least upper bound for $Y$. So if $B=\sup Y$ then $B-\epsilon$ is not an upper bound for $Y$, since $B-\epsilon<B$ and $B$ is the least upper bound.

And saying that $B-\epsilon$ is not an upper bound for $Y$ means exactly that there exists $y\in Y$ with $y>B-\epsilon$.

Answer 2

The only way that we can say that $\omega$ is not the supremum of a set $A$ means that there is an element $a_0 \in A$ that is not bounded by $\omega$, i.e., $\omega <a_0$.

Also, we know that if $\epsilon>0$, it follows that $\omega-\epsilon$ is smaller than $\omega$ and $\omega$ is the LEAST UPPER bound, so $\omega-\epsilon$ is not the supremum.

NOW, FOR THE QUESTION THAT YOU ARE TRYING TO SOLVE;

HINT:

If $\gamma \in A \cap B$, then $\gamma \in A$ and $\gamma \in B$.

Then $\gamma \leq lub A$ and $\gamma \leq lub B$ so $\gamma \leq max\{lubA,lubB \}$.

What if $max\{lubA,lubB\}$ is not the supremum?