# 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 [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]