Difference between revisions of "Passing to the infimum"
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{{Requires references|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}} | {{Requires references|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== | ==Statement== | ||
Latest revision as of 21:27, 19 April 2016
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Tidy up proof
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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:
- [ilmath]\forall a\in A[\text{inf}(A)\le a][/ilmath]
- [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
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