How to prove that infimum of $A = \{\frac{n}{2^n} : n \in N\}$ is 0?

Question

Could you explain this to me step by step?

I've mistaken, now it's correct. – Elise Svart Oct 23 '19 at 17:12 — Elise Svart, Oct 23 '19 at 17:12
See the corollary of the least upper bound property. – jl00 Oct 23 '19 at 17:15 — jl00, Oct 23 '19 at 17:15

score 1 · Answer 1 · answered Oct 23 '19 at 17:13

1

This set definitely admits an infimum. The only sets without an infimum are the one that do not admit any lower bound. This one is bounded below by $0$ so an infimum exists.

answered Oct 23 '19 at 17:13

Étienne Batelier

175

The empty set admits a lower bound, but has no infimum. But for non-empty sets, you are right. – Arthur Oct 23 '19 at 17:17
Oh, I didn't noticed it's only for natural numbers, sorry. – Elise Svart Oct 23 '19 at 17:17

How to prove that infimum of $A = \{\frac{n}{2^n} : n \in N\}$ is 0?

1 Answers1