arbitrary join
Let $P$ be a poset and $A\subseteq P$. The join of $A$ is the supremum^{} of $A$, if it exists. It is denoted by
$$\bigvee A\mathit{\hspace{1em}\hspace{1em}}\text{or}\mathit{\hspace{1em}\hspace{1em}}\underset{i\in I}{\bigvee}{a}_{i}\mathit{\hspace{1em}\hspace{1em}}\text{or}\mathit{\hspace{1em}\hspace{1em}}\bigvee \{{a}_{i}\mid i\in I\},$$ 
if the elements of $A$ are indexed by a set $I$:
$$A=\{{a}_{i}\mid i\in I\}.$$ 
In other words, $\bigvee A=supA$, where the equality is directed in the sense that one side is defined iff the other side is, and when this is the case, both sides are equal. Dually, one defines the meet of $A$ to be the infimum^{} of $A$, if it exists. The meet of $A$ is denoted by $\bigwedge A$.
Remark. The concepts of $\bigvee $ and $sup$ of an ordered set are identical. Besides being notationally distinct, $\bigvee $ is often used in order theory, while $sup$ is more prevalent in analysis^{}. Moreover, $\bigvee $ is generally being viewed as a (partial) function on the powerset ${2}^{P}$ of the poset $P$, while $sup$ is frequently seen as an operation^{} on sequences (or more generally nets) of elements of $P$.

1.
As was remarked, given a poset $P$, let us view $\bigvee :{2}^{P}\to P$ as a partial function^{}. Then $\bigvee $ is defined for all singletons. In fact $\bigvee \{a\}=a$ for all $a\in P$. Dually, $\bigwedge \{a\}=a$.

2.
If $\bigvee $ is defined for all doubletons, then it is defined for all finite subsets of $P$. In this case, $P$ is called a joinsemilattice. Dually, $P$ is a meetsemilattice is $\bigwedge $ is defined for all doubletons.

3.
If $A\subseteq B$, then $\bigvee A\le \bigvee B$, provided that both joins exist. We also have a dual statement: $A\subseteq B$ implies that $\bigwedge B\le \bigwedge A$, provided that both meets exist.

4.
$\bigvee \mathrm{\varnothing}$ exists iff $P$ has a bottom $0$, and when this is the case, $\bigvee \mathrm{\varnothing}=0$. This is essentially the result of the previous bulleted statement. Dually, $P$ has a top $1$ iff $\bigwedge \mathrm{\varnothing}$ exists, and when this is the case $\bigwedge \mathrm{\varnothing}=1$.

5.
Simiarly $\bigvee P=1$ and $\bigwedge P=0$, where the equality is directed on both sides.

6.
It can be shown that if $\bigvee $ is a total function from ${2}^{P}$ to $P$, then $P$ is a complete lattice^{}. (see proof here (http://planetmath.org/CriteriaForAPosetToBeACompleteLattice)).

7.
Let $P$ be a poset such that $\bigvee $ is defined for all subsets (of $P$) of cardinality $\U0001d52a$, is it true that $\bigvee $ is defined for all subsets of cardinality $\le \U0001d52a$? The answer is no, even when $\U0001d52a$ is finite. A counterexample^{} can be constructed as follows.
$$\text{xymatrix}\mathrm{@}!=1pt\mathrm{\&}1\text{ar}\mathrm{@}.[d]\text{ar}\mathrm{@}[rrdd]\mathrm{\&}\mathrm{\&}\mathrm{\&}\text{ar}\mathrm{@}[rd]\text{ar}\mathrm{@}[ld]\mathrm{\&}\mathrm{\&}a\mathrm{\&}\mathrm{\&}b\mathrm{\&}c$$ Let $C$ be an infinite chain with a top element $1$ (this can be found by taking the set of natural numbers and dualize the usual order). Adjoin three elements $a,b,c$ to $C$ so that $a$ and $b$ are below all elements of $C$, and $c$ is covered by $1$, and no two of $a,b,c$ are comparable^{}. This new poset $P$ has the property that any three distinct elements have a join. For example, $\bigvee \{a,b,c\}=1$. However, $\bigvee \{a,b\}$ does not exist.
Title  arbitrary join 

Canonical name  ArbitraryJoin 
Date of creation  20130322 17:27:53 
Last modified on  20130322 17:27:53 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06A06 
Related topic  CompleteLattice 
Related topic  CompleteSemilattice 
Defines  arbitrary meet 
Defines  infinite join 
Defines  infinite meet 