prove lim inf sn=- lim sup-Sn for every sequence Sn

Since lim inf Sn is the largest limit point of the sequence (Sn), we have:

For any ε > 0, there exists N such that Sn < lim inf Sn + ε for all n ≥ N.

Similarly, lim sup (-Sn) is the smallest limit point of the sequence (-Sn), so we have:

For any ε > 0, there exists N such that -Sn > lim sup (-Sn) - ε for all n ≥ N.

Now, let ε > 0 be arbitrary. By the first property above, there exists N1 such that Sn < lim inf Sn + ε for all n ≥ N1. This implies that -Sn > -(lim inf Sn + ε) for all n ≥ N1.

Using the second property above, there exists N2 such that -Sn > lim sup (-Sn) - ε for all n ≥ N2. This implies that Sn < - (lim sup (-Sn) - ε) for all n ≥ N2.

Let N = max{N1, N2}. Then for all n ≥ N, we have:

Sn < lim inf Sn + ε ≤ - (lim sup (-Sn) - ε) < -Sn

Thus, we have Sn < -Sn for all n ≥ N, which means Sn = 0 for all n ≥ N. Therefore, lim inf Sn = lim sup (-Sn) = 0, which proves the desired result