The length of a poset Let $(X,\le)$ be a (nonempty) poset. The length of $(X,\le)$ is defined by
$$\operatorname{len}(X)=\sup\{|C|-1\mid C\subseteq X \text{ is a chain} \}$$
Can $\sup$ be replaced by $\max$? (without changing notion/definition of $\text{len}$)
 A: Using ordinals we can produce simple examples: Consider for instance the set $\bigcup_n\omega_n\times\{n\}$, where $(\alpha,n)<(\beta,m)$ iff $n=m$ and $\alpha<\beta$. Here, there are chains of any length below $\aleph_\omega$, but not of length $\aleph_\omega$. (This example also shows that the cofinality of a partially ordered set can be singular.)
In the same spirit but not requiring any use of ordinals, consider $\bigcup_n n\times\{n\}$, with the same ordering. Here we have chains of any finite length, but no infinite chains. 
Or, for any infinite cardinal $\kappa$, consider $\bigcup_{\alpha<\kappa}\alpha\times\{\alpha\}$, again with the same ordering. We have chains of every ordinal length below $\kappa$, but none of length $\kappa$ itself. If $\kappa$ is a limit cardinal, this gives us another example as in the first paragraph: Chains of every cardinality below $\kappa$, so the supremum of their lengths is $\kappa$, and yet no chain of length $\kappa$ itself.
A: $\sup$ can always be replaced by $\max$, if you're dealing with finite collections.  However $\sup \{1,2,3,\ldots\}=\infty$ is defined, while $\max\{1,2,3,\ldots\}$ isn't. j
Also $\sup\{-1,-\frac{1}{2},-\frac{1}{4},-\frac{1}{8},\ldots\}=0$ while $\max \{-1,-\frac{1}{2},-\frac{1}{4},-\frac{1}{8},\ldots\}$ is again undefined.  That's not really relevant to this problem, though, since your elements are isolated.
