Question

Let A and B be non-empty, bounded subsets of R. (a) Why does sup(AUB) exist? (b) Prove that sup(AUB) = max{sup A, sup B}.

Answer 1

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Explanation:

Answer 2

Answer with Step-by-step explanation:

Let A and B be non- empty bounded subset of R

a.We have to find why
`sup(A\cup B)`exist

If A and B are bounded set

Then there exist constant such that

`a\leq A\leq b` and
`c\leq B\leq d`

Then , sup of A =b and sup of B=d

When a set is bounded then all elements lie in the set are lie between the constants s and t.

All elements are less than or equal to t then t is supremum of set.

Because both set are bounded and sup of both set A and B are exist.All elements A union B are less than or equal to sup A or sup B.

`sup(A\cup B)=max(sup A, sup B)`

Then,
`sup (A\cup B)` exist.

b.We have to prove that

`sup (A\cup B)=max(sup A,sup B)`

Suppose ,A =(1,2) and B=(2,3)

Sup A=2 , sup B=3

`(A\cup B)=(1,2)\cup (2,3)`

Upper bound of
`A\cup B)=3`

Hence,
`Sup (A\cup B)=3`

If A=(4,5),B=(2,3)

Sup A=5,Sup B=3

`A\cup B=(4,5)\cup (2,3)`

`Sup(A\cup B)=5`

Hence,
`Sup(A\cup B)=5`

Hence, we can say that
`sup(A\cup B)=max(sup A,sup B)`.