connected posetConnectedPoset2007-05-23 08:27:562007-05-23 14:15:43Definitionconnectedconnected component\usepackage{amssymb,amscd}
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\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}}Let $P$ be a poset. Write $a\perp b$ if either $a\le b$ or $b\le a$. In other words, $a\perp b$ if $a$ and $b$ are comparable. A poset $P$ is said to be \emph{connected} if for every pair $a,b\in P$, there is a finite sequence $a=c_1, c_2,\ldots, c_n=b$, with each $c_i\in P$, such that $c_i\perp c_{i+1}$ for each $i=1,2,\ldots,n-1$.
For example, a poset with the property that any two elements are either bounded from above or bounded from below is a connected poset. In particular, every semilattice is connected. A fence is always connected. If $P$ has more than one element and contains an element that is both maximal and minimal, then it is not connected. A \emph{connected component} in a poset $P$ is a maximal connected subposet. In the last example, the maximal-minimal point is a component in $P$. Any poset can be written as a disjoint union of its components.
It is true that a poset is connected if its corresponding Hasse diagram is a connected graph. However, the converse is not true. Before we see an example of this, let us recall how to construct a Hasse diagram from a poset $P$. The diagram so constructed is going to be an undirected graph (since this is all we need in our discussion). Draw an edge between $a,b\in P$ if either $a$ covers $b$ or $b$ covers $a$. Let us denote this relation between $a$ and $b$ by $a \asymp b$. Let $E$ be the collection of all these edges. Then $G=(P,E)$ is a graph where elements of $P$ serve as vertices and $E$ as the constructed edges. From this construction, one sees that a finite path exists between $a,b\in V(G)=P$ if there is a finite sequence $a=d_0,d_1,\ldots, d_m=b$, with each $d_i\in V(G)$, such that $d_i\asymp d_{i+1}$ for $i=1,\ldots,m-1$. In other words, $a$ and $b$ can be ``joined'' by a finite number of edges, such that $a$ is a vertex on the first edge and $b$ is the vertex on the last edge.
As promised, here is an example of a connected poset whose underlying Hasse diagram is not connected. take the real line $\mathbb{R}$ with $\infty$ adjoined to the right (meaning every element $r\in \mathbb{R}$ is less than or equal to $\infty$). Then the resulting poset is connected, but its underlying Hasse diagram is not, since no element in $\mathbb{R}$ can be joined to $\infty$ by a finite path.