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

1. Aug 12, 2009

### fmam3

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

1. The problem statement, all variables and given/known data
Suppose that the sequence $$(s_n)$$ is bounded in $$\mathbb{R}$$. Prove: Given any open interval containing $$\limsup s_n$$, there exists infinitely many $$n$$ with $$s_n$$ inside that interval.

2. Relevant equations
In light of the given statement, what is the issue with "open interval"? That is, does this statement hold (or not hold) if it said "closed interval" instead? And why?

2. Aug 12, 2009

### slider142

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.

3. Aug 12, 2009

### Dick

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?

4. Aug 12, 2009

### fmam3

@Dick
I understand perfectly the $$(L - \varepsilon, L + \varepsilon)$$ 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 :)