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I was asked to show that if every subset of a poset has an infimum then every such subset has a supremum. I did my proof and now I realize that what I was calling "infimum" was actually "a smallest element in given subset" (i.e. minimal element), so I actually proved that if every subset has a minimal element it has a maximal element. But I don't see why we should require the original statement to be true.

I don't know how to draw things on here (seen a few questions with really pretty diagrams, 2nd question How do I draw graphs and things on this site?) Anyway my attempt at a counter-example to my original problem is four points in a square where the top two are not connected but are both connected to both of the bottom points and the bottom points are connected to each other but the top points are not connected to each other.

In particular, the two top points are a subset and either of the bottom points are an infimum. Well maybe this isn't a counterexample... I suppose the trick lies in understanding that the infimum is unique. Why must it be unique?

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    (http://mathworld.wolfram.com/CrownGraph.html)2012-09-22

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Let $\langle P,\le\rangle$ be a poset in which every subset has an infimum, and let $A\subseteq P$ be arbitrary. Let $U=\{p\in P:a\le p\text{ for all }a\in A\}$, the set of upper bounds of $A$, and let $u=\inf U$. Clearly each $a\in A$ is a lower bound for $U$, so $a\le u$ for each $a\in A$, and therefore $u$ is actually the minimum element of $U$. In other words, $u$ is the smallest upper bound for $A$, which is precisely what is meant by saying that $u=\sup A$. Thus, every subset of $P$ has a supremum.

Note that the emptyset is not treated any differently from the non-empty sets: $u=\inf\varnothing$ if and only if (1) $u\le p$ for all $p\in\varnothing$, which is vacuously true of all $u\in P$, and (2) if $v\in P$ also has property (1), then $v\le u$. Since every $u\in P$ has property (1), property (2) says that $u$ is the maximum element of $P$, usually denoted by $\top$ or $1_P$. Similarly, if $\varnothing$ has a supremum, it must be the minimum element of $P$, usually denoted by $\bot$ or $0_P$.

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    @dustanalysis: It has nothing to do with Zorn’s lemma. $U$ can’t be empty because because $P$ has a maximum element, $\inf\varnothing$, and that maximum element is definitely in $U$.2012-09-22