Thread: For any open interval containing limsup s_n, there exists infinitely many s_n.....

@Dick I understand perfectly the argument --- in fact, that was how I proved the original (unmodified) statement. But thanks for the input :) @slider142 Ahh... I see, so the only reason of using the phrase "open interval" is to avoid trivial cases where a closed interval is a singleton / contains only one point. Great! Thanks :)

Call your limsup L. The point to saying the interval is open is that you then can find an epsilon such that (L-epsilon,L+epsilon) is contained in the interval. Suppose that subinterval contains only finitely many sn's. Then can L really be the limsup?

A singleton is a closed interval which is not an infinite set, and thus can't contain an infinite set. Every non-singleton real interval contains an open interval, so the theorem holds for this more convoluted object. It is much easier to say open interval.