# Suprema question

• Jan 2nd 2011, 10:54 AM
worc3247
Suprema question
Let S,T be non-empty bounded subsets of the real numbers. Prove that $S \cup T$ is bounded above.
How would you start this question?
• Jan 2nd 2011, 11:37 AM
FernandoRevilla
Quote:

Originally Posted by worc3247
Let S,T be non-empty bounded subsets of the real numbers. Prove that $S \cup T$ is bounded above.
How would you start this question?

If $s\geq 0,\;t\geq 0$ are upper bounds for $S$ and $T$ respectively, prove that $x\leq s+t$ for all $x\in S\cup T$ .

Fernando Revilla
• Jan 2nd 2011, 11:45 AM
Kevin123
Thats not right, you should take max{s,t} as upper bound. For example think of S=T=[-2,-1] and s=t=-1. In this case s+t=-2 is no upper bound of the union.
• Jan 2nd 2011, 12:01 PM
FernandoRevilla
Quote:

Originally Posted by Kevin123
Thats not right. mFor example think of S=T=[-2,-1] and s=t=-1. In this case s+t=-2 is no upper bound of the union.[/

It is right, I said

$s\geq 0,\;t\geq 0$

upper bounds, not minimum upper bounds.

Quote:

you should take max{s,t}
Or

$\max \{s,t\}+\pi$

:)

Fernando Revilla
• Jan 2nd 2011, 12:07 PM
Kevin123
Oh sorry, I did not see that you assume s,t >= 0. Of course then your answer is right... :-)
• Jan 2nd 2011, 12:14 PM
worc3247
I think I can see the general idea, but am having trouble writing a formal proof. Here is what I have thus far:
Let $\alpha \geq 0$and $\beta \geq 0$ such that $\alpha , \beta$ are upper bounds for S and T respectively.
If we set $l =$max{ $\alpha , \beta$}, then:
$l \geq x, \forall x \in S \cup T$. Therefore $S \cup T$ is bounded above.

Is this along the right lines or am I way off?
• Jan 2nd 2011, 12:33 PM
Kevin123
You cannot assume that the upper bound is in S. Think of S=(0,1). Then any upper bound is greater or equal 1.
• Jan 2nd 2011, 12:39 PM
worc3247
Ok, so I need to set $\alpha$ and $\beta$ so that they are greater than the supremum of the sets?