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 A,B⊆X be subsets of X where (X,⪯) is a poset. Then:
- If ∀a∈A∃b∈B[b≤a] then inf(B)≤inf(A) (provided both infima exist and are comparable)
Proof
Suppose we have ∀a∈A∃b∈B[b≤a] and that inf(B)>inf(A) - we shall reach a contradiction.
- By the definition of the infimum:
- ∀a∈A[inf(A)≤a]
- ∀x∈X∃a∈A[x>inf(A)⟹a<x] - there is no greater "lower bound" that is actually a lower bound.
- Note that by hypothesis: ∀a∈A∃b∈B[inf(B)≤b≤a] this means ∀a∈A[inf(B)≤a]
This contradicts that inf(A) was the infimum of A as inf(B) is greater than inf(A) and a lower bound of A
References
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