Question

Show that if a sequence (xn) is unbounded below then (xn) has a

subsequence (xnj) with xnj < −j for all j ∈ N.

Answer 1

To prove that an unbounded below sequence has a subsequence that is less than the negative of each natural number, one constructs a subsequence by inductively finding terms from the original sequence that fulfill the requirement for each natural number.

Step-by-step explanation:

If a sequence (xn) is unbounded below, it means that for any given number, there is always a term in the sequence that is lower than that number. We are asked to show that such a sequence has a subsequence (xnj) with xnj < −j for all j ∈ N (where N is the set of natural numbers).

Since the sequence is unbounded below, for j = 1, there exists at least one term xn1 such that xn1 < -1. Similarly, for j = 2, there exists a term xn2 with n2 > n1 such that xn2 < -2. Continuing this process inductively, for each j in N, we can find a term xnj in the sequence (xn) with nj > n(j-1) so that xnj < -j. This gives us a subsequence (xnj) such that xnj < -j for all j ∈ N, thus proving the statement.