Let $L$ be a finite partially ordered set where the maximum size of an antichain is $n$. I need to prove that $L$ can be expressed as a union of $n$ subchains.
Here is my attempt:
$L$ must have at least $n$ elements. If it has $n$ elements then it is the union of each of these elements (which individually form chains). If it has $n + 1$ elements we can take $n$ elements which cannot be compared and make them each into their own chain. Take the left over element which must be able to be compared with at least one more element and add it to that element's chain.
Proceed by induction. If $n+k$ elements can be made into $n$ subchains we must show that $n+k+1$ elements must can be made into $n$ subchains. Because there are more than $n$ elements, there must be at least one "maximal element" (or "minimal element", similarly defined) which is greater than every element it can be compared to and that can be compared to at least one element. Remove this element. Fit the $n+k$ left over elements into $n$ chains and then the "maximal element" can be placed at the top of one of the $n$ chains.
Is my proof correct? If (and only if) my proof is correct, I am interested in knowing how this result can be extended to infinite partially ordered sets. My book mentions that it is possible with "transfinite methods" but has not described what these are.
If my proof is not correct, what is missing?
Gratis.
Your proof is incorrect. The problem is that you aren't guaranteed to be able to fit element $n+k+1$ into any of the existing chains. As an example, consider the set $\{2, 3, 4, 5, 6\}$, ordered by divisibility:
The maximum size of an antichain in this set is $3$. Suppose we build $3$ chains $2-6$, $3$, and $5$. We can't fit $4$ into any of these chains, so the induction fails.
The question has been edited several times in a way that invalidates existing answers. I'm leaving this up in a state that answers the original question, but it may no longer address the question's current state.