Regarding chains and antichains in a partially ordered infinite space

Question

I've been given this as an exercise.

If P is a partially ordered infinite space, there exists an infinite subset S of P that is either chain or antichain.

This exercise was given in the Axiom of Choice section of the class. I answered it using Zorn's lemma but I'm not 100% sure. Can you give me any hints?

I used Hausdorff's maximal principle and said that there would be a chain that is maximal. And I used Zorn's lemma on the partially ordered subspace of P that contains all the subsets of P that are antichains. But I don't know if these solutions are correct and even if they are I'm stuck at the infinite part.

I was informed that this question is missing context or other details. I'm sorry but that was exactly how it was given to me so I don't know how to correct it.

I used Hausdorff's maximal principle and said that there would be a chain that is maximal. And I used Zorn's lemma on the partially ordered subspace of P that contains all the subsets of P that are antichains. But I don't know if these solutions are correct and even if they are I'm stuck at the infinite part — Amontillado, Jun 08 '15 at 00:20
Well, the whole point is the infinite part, and neither Hausdorff's principle nor Zorn's lemma will give you that. — Andrés E. Caicedo, Jun 08 '15 at 00:28
In the maximal chain I said that if it is finite then there must be a bigger element in P that isn't in the chain and if that's the case then it won't be maximal but I don't know how I can be sure of the existence of such an element... — Amontillado, Jun 08 '15 at 00:34
Your Zorn's Lemma argument doesn't work. An infinite partially ordered set may have a finite maximal chain and a finite maximal antichain. — bof, Jun 08 '15 at 03:19
See Given an infinite poset, show that it contains either a infinite chain or an infinite totally unordered set and Prove that for any infinite poset there is an infinite subset which is either linearly ordered or antichain — Martin Sleziak, Jun 08 '15 at 04:23
When a question is put on hold, the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it is good to avoid minor edits and improve your question as much as possible with the next edit.) Which is what happened to your question after you edited it. — Martin Sleziak, Jun 08 '15 at 04:31
You wrote that you do not know how to add the missing context. You should have a look at a link provided in the above message: http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question#9960 (I copied text of some of your comments into the question, since one way to provide context is to show your own work.) — Martin Sleziak, Jun 08 '15 at 04:32
In any case, even if your question is reopened, it will probably be closed again, this time as a duplicate of one of the questions I linked above. — Martin Sleziak, Jun 08 '15 at 04:47
@Martin, thanks a lot for the links. It's what I was looking for. — Amontillado, Jun 08 '15 at 04:53

score 1 · Answer 1 · answered Jun 08 '15 at 03:06

1

HINT: The easiest argument is to use the infinite Ramsey theorem; you need just two colors, one for pairs that are related in $P$, and one for pairs that are incomparable in $P$. There is a fairly easy proof of the theorem at the link.

answered Jun 08 '15 at 03:06

Brian M. Scott

631,399

I'm not familiar with these concepts. This exercise was given in the Axiom of Choice section of the class. Is there a way using the axiom of choice? – Amontillado Jun 08 '15 at 03:42
@Amontillado: Yes, but the one that occurs to me is essentially just a special case of the induction step of the argument at the link. Specifically, assume that $r=1$ and $c=2$, and try to carry out that argument. – Brian M. Scott Jun 08 '15 at 03:46
Ok thanks, I'll give it a shot – Amontillado Jun 08 '15 at 03:49
@Amontillado: You’re welcome. I’ll be happy to answer questions if you get stuck, though I may be going offline for a few hours pretty soon. – Brian M. Scott Jun 08 '15 at 03:50
That's ok. I'll give it a rest for today anyway. I'll try again tomorrow – Amontillado Jun 08 '15 at 03:52

Regarding chains and antichains in a partially ordered infinite space

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