Let S be a nonempty set of real numbers that is bounded above, and let s be its least upper bound. Prove that there exists a sequence {a_n} such that each a_n belongs to S and s is the limit of the sequence {a_n}.
Added by Monique B.
Step 1
Since S is bounded above, there exists a real number M such that for every element x in S, x ≤ M. This means that S is bounded by M from above. Show more…
Show all steps
Close
Your feedback will help us improve your experience
Sri K and 80 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Provide a formal proof for this: "the limit inferior of a bounded sequence is a finite real number"
Adi S.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD