Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and(xmk ) of (xn) such thatlimk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.

Question

Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and(xmk ) of (xn) such thatlimk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.
🧐 Not the exact question you are looking for?Go ask a question

Solution 1

To prove this, we will use the definitions of limit superior (lim sup) and limit inferior (lim inf) of a sequence.

  1. Let's start with lim sup. By definition, lim sup of a sequence (xn) is the smallest number 'L' such that for any positive number ε, there exists a natural number N such that for all Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study prob
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solv

This problem has been solved

Similar Questions

Let (xn) be a bounded sequence in R. Show that there exist subsequences (xnk ) and(xmk ) of (xn) such thatlimk→∞ xnk = lim sup xn and limk→∞ xmk = lim inf xn.

Let (xn)n≥1 and (yn)n≥1 be bounded sequence. Prove thatlim infn→∞ (xn + yn) ≥ lim infn→∞ xn + lim infn→∞ yn

Let A and B be non-empty bounded subsets of R. DefineA − B := {a − b : a ∈ A, b ∈ B}.Prove thatsup(A − B) = sup A − inf B

Let A and B be non-empty bounded subsets of R. DefineA − B := {a − b : a ∈ A, b ∈ B} .Prove thatinf(A − B) = inf A − sup B

If (xn) is a bounded sequence in R then the following statements about a numberx∗ ∈ R are equivalent

1/3