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.