Let $P$ denote a poset. Then the width of $P$ is defined as the supremum of the set of all cardinal numbers $\kappa$ such that $P$ has an antichain of cardinality $\kappa$. By the quasiwidth of $P$, let us mean the infimum of the set of all cardinal numbers $\kappa$ such that $P$ can be covered by $\kappa$-many chains. It can be shown that:
Proposition. The width of a poset is less than or equal to its quasiwidth.
To see this, use:
Proposition. Consider a fixed poset. Let $A$ denote an antichain therein and $K$ denote a covering by chains. Then there is a surjective partial function $A \leftarrow K$ defined by assigning to each $k \in K$ the unique $a \in A$ such that $a \in k$, whenever such an $a$ exists.
It follows that $|A| \leq |K|$, whenever $A$ is an antichain and $K$ is a covering of the poset by chains.
Anyway, what I'd like to know is:
Question. Does there exist a poset whose quasiwidth strictly exceeds its width?
From the reference provided by Random Jack:
What I'm calling the quasiwidth is usually called the chain covering number, denoted $c(P)$.
The width is denoted $w(P)$.
Its a general principle that $w(P) \leq c(P)$.
If $P$ is finite, then $w(P) = c(P)$.
If $\kappa$ is an infinite cardinal, then: $$w(\kappa \times \kappa) = \aleph_0, \qquad c(\kappa \times \kappa) = \kappa$$
Hence if $\kappa$ is uncountable, then $$w(\kappa \times \kappa) < c(\kappa \times \kappa).$$