Talk:Dilworth's theorem

History[edit]

Look at this.Kope (talk) 15:26, 8 March 2009 (UTC)[reply]

Locally finite posets?[edit]

Does anyone know if Dilworth's Theorem holds for locally finite posets? I'm having a hard time digging up a reference one way or another. (The Peles counterexample is not locally finite.) Vince Vatter (talk) 18:25, 6 June 2009 (UTC)[reply]

Examples?[edit]

Seriously, someone went through all the trouble of posting this without posting a single example? Ridiculous. —Preceding unsigned comment added by 66.49.227.118 (talk) 18:13, 25 January 2011 (UTC)[reply]

There are three examples in the article already: the one in the figure, the one involving Boolean lattices, and the one involving divisibility. They are not in a section with a big heading "EXAMPLES" but that doesn't make them non-examples. On the other hand, maybe they could use to be spelled out in more detail. —David Eppstein (talk) 18:57, 25 January 2011 (UTC)[reply]

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Chain decompositions[edit]

I've noticed "chain decomposition" redirects here, is there any good reason it does not have its own page? I don't the concept is so well defined here. LudwikSzymonJaniuk (talk) 12:40, 22 February 2019 (UTC)[reply]

@LudwikSzymonJaniuk: Maybe you mean Heavy path decomposition? wumbolo ^^^ 14:23, 22 February 2019 (UTC)[reply]

That's related, but not the same. In the context of this theorem, a chain decomposition would be a partition of a partial order into total orders. The theorem states how many total orders you need. I'm not sure, is there much to say about chain decompositions beyond that? —David Eppstein (talk) 17:21, 22 February 2019 (UTC)[reply]

I just came here trying to understand what the definition of chain decomposition is, and I didn't find it. LudwikSzymonJaniuk (talk) 20:33, 25 February 2019 (UTC)[reply]

The definition is actually right there in the first sentence of the article. It's just not stated that it's the definition of that term. —David Eppstein (talk) 21:40, 25 February 2019 (UTC)[reply]

Proof using König's theorem[edit]

The article says "Let A be the set of elements of S that do not correspond to any vertex in C; then A has at least n - m elements". So how does it prove that the size of the biggest antichain and of the smallest partition into chains is equal? To me there only seems to be a lower bound on the antichain size. Llama carl (talk) 17:44, 16 March 2019 (UTC)[reply]

The matching upper bound is very very easy. So proving the lower bound is the main part. —David Eppstein (talk) 17:47, 16 March 2019 (UTC)[reply]