Question

Let S be a set of real numbers. Then S has
a supremum if s has
O a lower bound belongs to S
O b. an upper bound

O c. upper bound belongs to S
d. A lower bound​

Answer 1

Answer:

This axiom states that any non-empty set S ⊂ R that is bounded above has a supremum; in other words, if S is a non-empty set of real numbers that is bounded above, there exists a b ∈ R such that b = sup S

Answer 2

Concept:

This Question can be solved by using the concept of set theory.

Set theory is the mathematical theory of sets (well-defined collections), or elements, of the set.

Given:

A set S of real numbers.

To find:

The condition under which the set S has a Supremum

Solution:

In terms of mathematics, The supremum of a set S ⊆ R which is bounded above is an upper bound b ∈ R of S such that b ≤ u for any upper bound u of S, denoted by b = supS

In terms of sets,  supremum of a set is the least upper bound of the set. Supremum does not have to belong to the set.

For  Example: the set S= { $\frac{-1}{n}$ | for all natural values of n}. The supremum is 0 but 0 is not part of the set.

Since, Any nonempty set of real numbers which is bounded above has a supremum

Hence, the given set S has a supremum if it has an upper bound

So, option (b) is correct choice