Passing to the infimum

From Maths
Jump to: navigation, search
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Tidy up proof
(Unknown grade)
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
I've searched and searched and I've found passing to the infimum used but never actually stated! This is what I think the theorem states, however as a proof is presented of the statement, the statement is at least correct

Statement

Let [ilmath]A,B\subseteq X[/ilmath] be subsets of [ilmath]X[/ilmath] where [ilmath](X,\preceq)[/ilmath] is a poset. Then:

  • If [ilmath]\forall a\in A\exists b\in B[b\le a][/ilmath] then [ilmath]\text{inf}(B)\le\text{inf}(A)[/ilmath] (provided both infima exist and are comparable)

Proof

Suppose we have [ilmath]\forall a\in A\exists b\in B[b\le a][/ilmath] and that [ilmath]\text{inf}(B)>\text{inf}(A)[/ilmath] - we shall reach a contradiction.

  • By the definition of the infimum:
    1. [ilmath]\forall a\in A[\text{inf}(A)\le a][/ilmath]
    2. [ilmath]\forall x\in X\exists a\in A[x>\text{inf}(A)\implies a<x][/ilmath] - there is no greater "lower bound" that is actually a lower bound.
  • Note that by hypothesis: [ilmath]\forall a\in A\exists b\in B[\text{inf}(B)\le b\le a][/ilmath] this means [ilmath]\forall a\in A[\text{inf}(B)\le a][/ilmath]

This contradicts that [ilmath]\text{inf}(A)[/ilmath] was the infimum of [ilmath]A[/ilmath] as [ilmath]\text{inf}(B)[/ilmath] is greater than [ilmath]\text{inf}(A)[/ilmath] and a lower bound of [ilmath]A[/ilmath]

References