Let S,T be non-empty bounded subsets of the real numbers. Prove that $\displaystyle S \cup T$ is bounded above.
How would you start this question?
If $\displaystyle s\geq 0,\;t\geq 0$ are upper bounds for $\displaystyle S$ and $\displaystyle T$ respectively, prove that $\displaystyle x\leq s+t$ for all $\displaystyle x\in S\cup T$ .
Fernando Revilla
It is right, I said
$\displaystyle s\geq 0,\;t\geq 0$
upper bounds, not minimum upper bounds.
Oryou should take max{s,t}
$\displaystyle \max \{s,t\}+\pi$
Fernando Revilla
I think I can see the general idea, but am having trouble writing a formal proof. Here is what I have thus far:
Let $\displaystyle \alpha \geq 0$and $\displaystyle \beta \geq 0$ such that $\displaystyle \alpha , \beta$ are upper bounds for S and T respectively.
If we set $\displaystyle l = $max{$\displaystyle \alpha , \beta$}, then:
$\displaystyle l \geq x, \forall x \in S \cup T$. Therefore $\displaystyle S \cup T$ is bounded above.
Is this along the right lines or am I way off?