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?

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- Jan 2nd 2011, 09:54 AMworc3247Suprema question
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? - Jan 2nd 2011, 10:37 AMFernandoRevilla

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 - Jan 2nd 2011, 10:45 AMKevin123
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.

- Jan 2nd 2011, 11:01 AMFernandoRevilla
It is right, I said

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

upper bounds, not minimum upper bounds.

Quote:

you should take max{s,t}

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

:)

Fernando Revilla - Jan 2nd 2011, 11:07 AMKevin123
Oh sorry, I did not see that you assume s,t >= 0. Of course then your answer is right... :-)

- Jan 2nd 2011, 11:14 AMworc3247
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? - Jan 2nd 2011, 11:33 AMKevin123
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, 11:39 AMworc3247
Ok, so I need to set $\displaystyle \alpha$ and $\displaystyle \beta$ so that they are greater than the supremum of the sets?