Question #3986

let S subset R be nonempty. show that if u= supremum S, then for every number n element N the number (u-1)/n is not an upper bound of S, but the number (u+1)/n is an upper bound of S.

Expert's answer

1) Let's prove the first part.

Assume that there is some n element N such that (u-1/n) is an upper bound of S.

By the definition supremum is the lowest upper bound. But easy to see that (u-1/n) < u and thus u isn't the lowest bound.

So it's not true that there is some n element N such that (u-1/n) is an upper bound of S.

It means that for every number n element N the number (u-1/n) is not an upper bound of S.

2) Let's prove the second part.

For any x element S it's true that x <= u, because u = sup S.

At the same time for any n element N it's true that u < u+1/n

So we have for any x element S: x <= u < u+1/n. From the transitivity of real numbers x < u+1/n

Thus for any n element N (u+1/n) is greater than any element of S, which means that (u+1/n) is an upper bound.

Assume that there is some n element N such that (u-1/n) is an upper bound of S.

By the definition supremum is the lowest upper bound. But easy to see that (u-1/n) < u and thus u isn't the lowest bound.

So it's not true that there is some n element N such that (u-1/n) is an upper bound of S.

It means that for every number n element N the number (u-1/n) is not an upper bound of S.

2) Let's prove the second part.

For any x element S it's true that x <= u, because u = sup S.

At the same time for any n element N it's true that u < u+1/n

So we have for any x element S: x <= u < u+1/n. From the transitivity of real numbers x < u+1/n

Thus for any n element N (u+1/n) is greater than any element of S, which means that (u+1/n) is an upper bound.

## Comments

Assignment Expert21.02.14, 13:41Dear manish,

our proof is complete. What do you want to be done?

manish03.02.14, 18:03prove converse of this .

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