82.4k views
3 votes
Prove that if a real valued sequence (aₙ​) does not have a lower bound then there is a subsequence (aₙₖ) that converges to negative infinity in the extended reals.

User Tahesha
by
9.1k points

1 Answer

6 votes

Final answer:

A subsequence that converges to negative infinity can be constructed from a real valued sequence without a lower bound by choosing sequence members such that each subsequent member is below every negative integer threshold.

Step-by-step explanation:

To prove that if a real valued sequence (a_n) does not have a lower bound then there is a subsequence (a_nk) that converges to negative infinity in the extended reals, we can use a constructive approach to define such a subsequence. Since (a_n) is unbounded below, for each natural number k, there exists an index n_k such that a_nk < -k. We pick n_1 first such that a_n1 < -1, then find n_2 > n_1 such that a_n2 < -2, and continue this process inductively. By construction, the subsequence (a_nk) satisfies that a_nk < -k, and thus, as k approaches infinity, (a_nk) converges to negative infinity.

The fact that the sequence does not have a lower bound means there is always a sequence member lower than any proposed bound; hence we can always find a suitable element for our subsequence that satisfies the condition to make the subsequence approach negative infinity.

User Nathan Bierema
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories