# Clarification on part of answer surrounding limsup, liminf

I am reading Paramanand Singh's very excellent solution here:

However, I am confused on the last part. Could anybody help me clarify it?

What do we gather from the above discussion? Well we have the following two obvious properties for $$M = \limsup a_{n}$$:

1. Any number greater than $$M$$ is superior so if $$\epsilon > 0$$ then $$a_{n} \leq M + \epsilon$$ for all large values of $$n$$.
2. Any number less than $$M$$ is not superior so if $$\epsilon > 0$$ then $$a_{n} > M - \epsilon$$ for infinitely many values of $$n$$.
• With all due respect to the previous discussion, I feel strongly the best definition of $\limsup a_n$ is as the largest subsequential limit of $a_n.$
– zhw.
Aug 17 '16 at 22:10
• you should have asked this as a comment in my answer to linked question so that everything is put at one place. Anyway I will add some details in my answer to add more clarity. Aug 18 '16 at 8:11
• @zhw.: Agree! the largest subsequential limit is very useful in applications. But I have never resorted in general to subsequences. Thus Bolzano Weierstrass for me is that "any bounded infinite set has an accumulation point" instead of "any bounded sequence has a convergent subsequence". Perhaps its more a matter of taste and preference. Aug 18 '16 at 8:14

By the definition of a limit and the lim sup, if $S_n$ denotes the supremum of all $a_k$ for $k\ge n$, for any $\epsilon > 0$ there exists an $m$ such that for all $n>m$ we have $|S_n-M|< \epsilon$, and this means that $S_n\le M+\epsilon$ for all $n>m$, which in turn implies that $a_n\le M+\epsilon$. This is the first assertion. It also implies $S_n> M-\epsilon$ for all $n>m$. This must mean that for any $n>m$ there exists a $k\ge n$ with $a_k\ge M-\epsilon$, for otherwise we would have $S_n\le M-\epsilon$. Since for any given natural $n$ we can find a $k\ge m$ such that $a_k\ge M-\epsilon$, there must be infinitely many $a_k$ which satisfy that property, so we get the second assertion.