Passing to the infimum

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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 $A,B\subseteq X$ be subsets of $X$ where $(X,\preceq)$ is a poset. Then:

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

Proof

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

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

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