set theory sup of a union

**CarmineCortez** · October 21st 2008, 11:14 AM

I have an assignment quesiton :

" let A_i be a subset of the reals and i is element of I = (1,...,n)

now let A = UNION (i is element of I) A_i

show that sup(A) = sup_iEI(sup(A))"

I hope that is understandable...

so taking the limit of both sides I have limA = lim U A_i, I'm not sure what the limit of the union of a set is. Am I on the right track?

**Plato** · October 21st 2008, 01:03 PM

Let $I = \left\{ {1,2, \cdots ,n} \right\}$ , I take this to be your question.
$\sup (A) = \sup \left( {\bigcup\limits_{i \in I} {A_i } } \right) = \mathop {\sup }\limits_{i \in I} \left[ {\sup \left( {A_i } \right)} \right]$ .
For notation: $\alpha _i = \sup \left( {A_i } \right)\,,\,\alpha = \sup \left\{ {\alpha _1 ,\alpha _2 \cdots ,\alpha _n } \right\}\,\& \,\alpha ' = \sup (A)$
$\left( {\forall x \in A} \right)\left( {\exists m \in I} \right)\left[ {x \in A_m \Rightarrow \,x \leqslant \alpha _m \leqslant \alpha } \right]$ but that means that $\alpha ' \leqslant \alpha$ (WHY?)

Now suppose that $\alpha ' < \alpha$ and find a contradiction to prove that $<br /> \alpha ' = \alpha$ .

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