Suprema question

**worc3247** · January 2nd 2011, 09:54 AM

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?

**FernandoRevilla** · January 2nd 2011, 10:37 AM

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

**Kevin123** · January 2nd 2011, 10:45 AM

That`s 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.

**FernandoRevilla** · January 2nd 2011, 11:01 AM

Originally Posted by Kevin123

That`s 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.

you should take max{s,t}

Or

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

Fernando Revilla

**Kevin123** · January 2nd 2011, 11:07 AM

Oh sorry, I did not see that you assume s,t >= 0. Of course then your answer is right... :-)

**worc3247** · January 2nd 2011, 11:14 AM

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?

**Kevin123** · January 2nd 2011, 11:33 AM

You cannot assume that the upper bound is in S. Think of S=(0,1). Then any upper bound is greater or equal 1.

**worc3247** · January 2nd 2011, 11:39 AM

Ok, so I need to set $\alpha$ and $\beta$ so that they are greater than the supremum of the sets?

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