Equivalent formulations of completeness

[math771]

olarBear: Yes, all of these formulations of completeness are equivalent. It is a fundamental property of real numbers and metric spaces, so it makes sense that any formulation of completeness would be equivalent to the others.

To answer your specific questions, 3 does imply 1, and 5 and 6 do become equivalent statements when considering ℝ as a metric space with the standard Euclidean metric. This is because the standard Euclidean metric satisfies the conditions for a metric space to have a Cauchy completeness and nested interval property.

I hope this helps! Good luck with your proofs.

 

[Bipolarity]

So far, I have encountered the following formulations of completeness and was wondering whether they are all equivalent:

1) Supremum principle: Every nonempty bounded subset of ℝ has a supremum.
2) Infimum principle: Every nonempty bounded subset of ℝ has an infimum.
3) Monotone sequence property: Every monotonic bounded sequence in ℝ converges to an element of ℝ
4) Dedekind completeness: If S and T form a Dedekind cut of ℝ such that S < T, then either S has a largest element or T has a smallest element.
5) Cauchy completeness: Every Cauchy sequence in a metric space converges to an element of that metric space.
6) Nested interval property: The (infinite) intersection of every nested sequence of closed balls (in an arbitrary metric space) whose radii tend to zero is nonempty.

I have proved equivalence of 1, 2 and 4. I have also proved that 1 implies 3. Is it true that 3 implies 1? Clearly, 5 and 6 are defined for arbitrary metric spaces, but do they become equivalent statements to the others if we regard ℝ as a metric space with the standard Euclidean metric?

Thanks for clarifications. Do not state proofs. I just want to know whether they're equivalent. I will do proofs myself, thanks!

BiP

 

[fresh_42]

I would try to prove ##4 \Longleftrightarrow 6##. Since ##5## and ##4## should be equivalent, only ##3 \Longrightarrow n## will be left with any ##n\neq 3##. It's probably easiest to prove ##n =1## or ##n=2##.