Question

Prove carefully that if A = { 1/n : n ∈ N}, then inf(A) = 0.

Answer 1

To prove that the infimum of the set A = { 1/n : n ∈ N} is 0, we need to show two things: (1) The number 0 is a lower bound for A. (2) Any number greater than 0 is not a lower bound for A.

Step-by-step explanation:

To prove that the infimum of the set A = { 1/n : n ∈ N} is 0, we need to show two things:

1. The number 0 is a lower bound for A.
2. Any number greater than 0 is not a lower bound for A.

Let's start with the first part. Since n ∈ N, the smallest value for n is 1. Hence, the smallest value in the set A is 1/1 = 1. This implies that 0 is a lower bound for A.

Now, let's tackle the second part. Suppose there exists a number k > 0 such that k is a lower bound for A. Since k is greater than 0, there exists a natural number m such that 1/m < k. But this implies that 1/(m+1) < 1/m < k. This contradicts the assumption that k is a lower bound for A because we found a smaller number in A.

Therefore, 0 is the greatest lower bound for A, which means the infimum of A is 0.