I was thinking about the equivalence of the two following definition of Limsup of a sequence. I find the definition 1 much more intuitive and I have been trying to convince myself of the equivalence of the two definitions given below but I haven't succeeded yet
Def 1 Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers and we define $S$ as the set of all subsequential limits of the sequence $(x_n)_{n\in \mathbb{N}}$. Then we define $\limsup x_n=\sup S$ Now this makes sense as if the sequence $(x_n)_{n\in \mathbb{N}}$ does not converge we can think about the supremum( and infimum) of its sub-sequential limits
Def 2: $\limsup x_n=\inf_{k}\{\sup {x_n}\}$ where the supremum over $n \geq k$
In this definition I see why it is defined the way it is, as we need the supremum of those $x_n$ as $n \to \infty$ so we take the infimum over the set of supremums (of a sets with fewer element as n increases we eventually are talking about the supremums of those $x_n$ with inifinitely high values of n . But I am not at all convinced that these two definitions are equivalent. Can someone help me see wy they are equivalent.This has troubled me and makes me uncomfortable everytime I have to deal with Limsups and Liminfs. Thanks
Just prove this carefully. Let $A = \inf_N \sup _{n>N} x_n, B = \sup S$.
Let $r>0$. You can find a convergent subsequence $x'$ such as $$ x'_n \to L > B - r $$ Now let $N>0$. $\sup _{n>N} x_n \ge L$ because infinitely many terms of $x'$ are in the set $\{x_n : n> N \}$. Hence $$A \ge L > B-r $$and as this is true for every $r>0$: $$A\ge B$$
Let $r>0$. There is a $N$ such as $$\sup _{n>N} x_n < A + r $$ Then for every convergent subsequence $x'$, as infinitely many terms are in $\{x_n : n> N \}$: $$ \lim x' < A+r $$and as it is true for every $x'$: $$ B \le A + r $$and as it is true for every $r>0$: $$ B \le A $$