0

In our Real Analysis textbook by Robert G. Bartle. I came across two different definitions of the Limit Superior ($LimSup$).

  1. If $x$ be the Limit Superior of the sequence $x_n$ then for every $\epsilon>0$ $$x_n<x+\epsilon$$ ultimately. And $$x_n>x-\epsilon$$ frequently.

  2. If $S$ be the set of all Subsequential Limits of $x_n$, then the Limit Superior of $x_n$ is defined as : $$Sup(S)$$

I cannot understand how to rigorously prove that the statement $1$ can imply statement $2$. Any ideas about how to approach the proof would be really helpful. Thanks in advance.

Uranium238
  • 31
  • This is almost a duplicate. Your 2 is the Rudin definition, where we work in the extended reals $[-\infty,+\infty]$ to ensure the non-emptysness of $S$ and the existence of its supremum. See also this. – Henno Brandsma Mar 13 '21 at 10:13
  • What Bartle Real Analysis textbook is this from? He certainly did not use either of these definitions in "The Elements of Real Analysis", 2nd Edition. Use "$x_n > x - \varepsilon$ frequently" to select a subsequence that converges to $x$, which means $x \in S$. Use "$x_n < x+\varepsilon$ ultimately" to show that for any $y > x$, there is no subsequence converging to $y$. – Paul Sinclair Mar 13 '21 at 18:34

0 Answers0