Hi,
i need to prove or disprove the following:
$\displaystyle Sup(A\cup B)=max\left \{ SupA, SupB \right \}$
thanks in advanced!
*edit:
both sets are bounded from above.
Hi,
i need to prove or disprove the following:
$\displaystyle Sup(A\cup B)=max\left \{ SupA, SupB \right \}$
thanks in advanced!
*edit:
both sets are bounded from above.
Prove (1) $\displaystyle \sup(A\cup B)\le\max(\sup A,\sup B)$ and (2) $\displaystyle \sup(A\cup B)\ge\max(\sup A,\sup B)$.
For (1), show that $\displaystyle \max(\sup A,\sup B)$ is an upper bound of $\displaystyle A\cup B$.
For (2), use the fact that $\displaystyle x\ge y$ and $\displaystyle x\ge z$ imply $\displaystyle x\ge\max(y,z)$.
I've found the following to be useful:
supremums of larger sets are larger and infimums of larger sets are smaller
Formally, (assuming the sets are bounded):
1. If $\displaystyle C\subseteq D$ then $\displaystyle sup(C)\leq sup(D)$
2. If $\displaystyle C\subseteq D$ then $\displaystyle inf(C)\geq inf(D)$
Proof of 1 - Let $\displaystyle u=sup(D)$. Let $\displaystyle x\in C$, then $\displaystyle x\in D$ and so $\displaystyle x\leq u$; i.e. $\displaystyle u$ is an upper bound of $\displaystyle C$. So $\displaystyle sup(C)\leq u$. The proof for infs is exactly similar.
For your original question, since $\displaystyle A\subseteq A\cup B$, $\displaystyle sup(A)\leq sup(A\cup B)$, and similarly for B.