Prove that the least upper bound of a set of a set of integers is

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Homework Help Overview

The discussion revolves around proving that the least upper bound of a set of integers is an integer. The original poster attempts to explore various methods of proof, including the well-ordering principle and the Archimedean property, while seeking alternative approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question the triviality of the well-ordering principle in this context and suggest that it may not lead to the supremum. Others propose proof by contradiction, considering the implications of the supremum not being an integer.
  • There are discussions about the definition of integers and the implications of the completeness axiom, with some participants expressing uncertainty about the converse of the axiom.
  • Several participants explore the concept of fractional parts and their relevance to the proof, with one participant expressing doubt about their initial reasoning.
  • The original poster seeks confirmation on whether the theorem can be proven without relying on the well-ordering principle.

Discussion Status

The discussion is active, with multiple lines of reasoning being explored. Participants are sharing their thoughts on various approaches and questioning the assumptions involved. There is no explicit consensus, but several productive ideas and clarifications have emerged.

Contextual Notes

It is assumed that the set in question is bounded above. Participants are also considering the implications of definitions and axioms related to real numbers and integers.

╔(σ_σ)╝
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Problem Statement:
Prove that the least upper bound of a set of integers is an integer.

Attempt:
Using well ordered principle this is very trivial. However, is there another way?

ANY comments or ideas relating to the topic would be highly appreciated.

It is assumed that the set mentioned is bounded above.

The mobile version of PF has bugs so the code that was supposed to be pre-generated for this thread did not show up :-(
 
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I don't think this is trivial with the well ordering principal... You can look at all integers larger than your set and find a smallest one, but that will definitely not be the supremum. If you have a different approach I'd be interested in seeing it.

I think proof by contradiction is good here: Suppose the supremum is not an integer. This means that its fractional part is non-zero. Can you find a contradiction with that?
 
I tried that approach but I hit a wall. It depends on how you define integers.
1.9999999999999999...=2

In my approach I showed that there is a contradiction if sup(S) was not an integer by using the archimedean property, that is, I can always find a better upper bound.
 
hmm may i ask, completeness axiom say " every nonempty set of real number that bounded from above have a least upperbound" does this correct conversely??

because sometimes i see when a "definition" they define "if then" but i find it always true conversely
 
I know what you mean; I found in a lot of theorems if can be replaced with iff. However, that doesn't apply here. Regarding the converse...
For the completeness axiom it is the case that every set that has a least. upper bound must be bounded above.
 
╔(σ_σ)╝ said:
I tried that approach but I hit a wall. It depends on how you define integers.
1.9999999999999999...=2

The fractional part of 1.9999... is not .9999..., it's 0. The fractional part is literally defined to be x-[x] where [x] is the largest integer smaller than or equal to x. So in this case, [1.999...]=2, x-[x]=0 as expected and the fractional part is not dependent on how you represented the number
 
Office_Shredder said:
The fractional part of 1.9999... is not .9999..., it's 0. The fractional part is literally defined to be x-[x] where [x] is the largest integer smaller than or equal to x. So in this case, [1.999...]=2, x-[x]=0 as expected and the fractional part is not dependent on how you represented the number
I guess I was overcomplicating things then; it is possible to do it that way. In fact, that was my original solution but I did not trust it.

The way I did it also works, right ?

Anyway the whole point of posting this question was to find another way of proving this theorem without well ordering.

So I must ask can this be done without well ordering ?
 
Does anyone have any other input?
 

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