Proving Suprema & Infima: Showing Sub B is an Upper Bound of A

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Homework Help Overview

The discussion revolves around the concept of supremum and infimum in the context of set theory, specifically addressing the relationship between two sets A and B. The original poster seeks to demonstrate that if the supremum of set A is less than the supremum of set B, then there exists an element in set B that serves as an upper bound for set A.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore proof techniques, including proof by contradiction, and discuss the existence of an element in B that satisfies the upper bound condition. Questions arise regarding the implications of sup B not being an element of B and whether a contradiction can be established under such circumstances.

Discussion Status

The discussion is ongoing, with participants sharing different perspectives on the problem. Some suggest exploring specific examples to gain insight, while others express uncertainty about the reasoning presented. There is no explicit consensus yet, but various lines of reasoning are being examined.

Contextual Notes

Participants are considering the implications of the definitions of supremum and the conditions under which the elements of sets A and B are defined. The discussion acknowledges the challenge posed by the case where sup B is not included in B.

lokisapocalypse
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I need to show that if sup A < sup B, then there is an element in B that is an upper bound of A.

Well I know that if sup B is in B, then sub B is the element in B that is an upper bound of A. But I don't know how to show this is true if sup B is not in B.
 
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You try a proof by contradiction?
 
You need to prove simply that [itex]\exists b \in B[/itex] such that [itex]\sup A \leq b \leq \sup B[/itex]. Assume not: then [itex]\sup A[/itex] is an upper bound of B, which contradicts the fact that [itex]\sup A < \sup B[/itex], and you're done.
 
AKG--

I'm not sure I understand your reasoning. What you say makes sense but that doesn't prove that such a b exists. Or does it and I just don't get it which is equally as possible.
 
Suppose b doesn't exist. Can you prove a contradiction? (such as sup A = sup B)
 
But what about the case where sup B is not in B? Then I don't think I can get that contradiction.
 
Then, try to find two sets A and B where b doesn't exist, but sup A < sup B!

We're lucky, we already know this will be impossible, but it still might give you insight.

(As always, it often helps to write out what the definitions say too)
 
Okay solid. Thanks.
 

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