# Are these two subsets of a poset already named?

In a partially ordered set $(X,≤)$,

• an upper set of a partially ordered set $(X,≤)$ is a subset $U$ with the property that, if $x \in U$ and $x≤y$, then $y \in U$.

• The dual notion is lower set, which is a subset $L$ with the property that, if $x \in L$ and $y≤x$, then $y \in L$.

I was wondering if the following two related concepts have been named already:

• a subset $S$ with the property that, if $x \in S$, then there exists a $y \in S, y \neq x$ s.t. $x≤y$.

• a subset $S$ with the property that, if $x \in S$, then there exists a $y \in S, y \neq x$ s.t. $y≤x$.

Thanks and regards!

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Every subset has these properties (take $y = x$). –  Qiaochu Yuan Jan 4 '13 at 7:41
If you make them strict inequalities you might use the made-up words 'maximumless' and 'minimumless'. –  anon Jan 4 '13 at 7:43
@QiaochuYuan: I should have required $y \neq x$. –  Tim Jan 4 '13 at 7:55
A union of infinite ascending chains. –  mjqxxxx Jan 4 '13 at 9:02

An infinite ascending chain (of $(X,\le)$) is a totally ordered subset $\{x_1,x_2,\ldots\}$ such that $x_1 < x_2 < \ldots$ Such a chain satisfies your first condition, as does any union of such chains. (To find the $y \ge x$ required, find a chain containing $x$ and take $y$ to be its successor.) On the other hand, let $S$ be a subset satisfying your first condition. For each $x\in S$, let $C_{x} \subseteq X$ be an infinite ascending chain beginning at $x$; the existence of at least one such chain for each $x \in S$ is guaranteed by your condition, and clearly $S=\bigcup_{x\in S}C_x$. The same reasoning applies to your second concept, with "ascending" replaced by "descending". We have shown the implication in both directions:
Given a subset $S$ of a partially ordered set $(X,\le)$, the following are equivalent:
1. $S$ is a union of infinite ascending (resp., descending) chains.
2. For each $x\in S$, there exists $y\in S$ such that $y>x$ (resp., $y<x$).