Question

Let A be a nonempty set of real numbers that is bounded above, let β = sup A, and suppose that β is not in A. Prove that for each ε > 0, the set (β - ε, β] contains infinitely many points of A.

Answer 1

To prove that for each ε > 0, the set (β - ε, β] contains infinitely many points of A, we can use the fact that β = sup A and that β is not in A. By contradiction, assuming (β - ε, β] contains only finitely many points leads to a contradiction with the definition of β as the least upper bound of A.

Step-by-step explanation:

Let A be a nonempty set of real numbers that is bounded above and β = sup A. Suppose β is not in A. We want to prove that for each ε > 0, the set (β - ε, β] contains infinitely many points of A.

Since β = sup A, for every positive ε, there exists an element a in A such that a > β - ε. If (β - ε, β] contains only finitely many points of A, then there must be a largest element x in (β - ε, β] that is an element of A. But this contradicts the fact that β is the least upper bound of A. Therefore, (β - ε, β] contains infinitely many points of A.