View source for 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}}
==Statement==
Let {{M|A,B\subseteq X}} be [[subset|subsets]] of {{M|X}} where {{M|(X,\preceq)}} is a [[poset]]. Then:
* If {{M|\forall a\in A\exists b\in B[b\le a]}} then {{M|\text{inf}(B)\le\text{inf}(A)}} (provided both [[infimum|infima]] exist and are [[comparable]])

==Proof==
Suppose we have {{M|\forall a\in A\exists b\in B[b\le a]}} and that {{M|\text{inf}(B)>\text{inf}(A)}} - we shall reach a contradiction.
* By the definition of the [[infimum]]:
*# {{M|\forall a\in A[\text{inf}(A)\le a]}} 
*# {{M|\forall x\in X\exists a\in A[x>\text{inf}(A)\implies a<x]}} - there is no greater "[[lower bound]]" that is actually a lower bound.
* Note that by hypothesis: {{M|\forall a\in A\exists b\in B[\text{inf}(B)\le b\le a]}} this means {{M|\forall a\in A[\text{inf}(B)\le a]}}
This contradicts that {{M|\text{inf}(A)}} was the [[infimum]] of {{M|A}} as {{M|\text{inf}(B)}} is greater than {{M|\text{inf}(A)}} and a [[lower bound]] of {{M|A}}

==References==
<references/>
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