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Suppose $P$ is a poset such that there exist (strong) antichains of size $n$ for all $n \in {\bf N}$; i.e. there exist sets $S_n$ of size $n$ in $P$ such that no pair of elements of $S_n$ has a common lower bound.

Must $P$ have an infinite antichain?

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    Posts that merely state a question are discouraged. You can improve your post with additional context. Where did you encounter the problem? Why is it of interest? What background do you have as you approach the problem? At the moment, there is nothing in this post to distinguish it from a homework problem directly copied to this site. You can edit the post to improve it. – 2017-02-20
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    While the question indeed lacks context, it is clear and sounds interesting to me. I have seen many questions like this on this website and probably have been guilty myself of asking some. I have voted to reopen it. – 2017-02-20

2 Answers 2

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If there is an infinite set of minimal elements (or rather elements that below them the order is linear), we're done.

So let's work under the assumption there are no minimal elements (or rather, every element has two incompatible smaller elements), as there are only finitely many of them, and they have to be in every maximal antichain.

Now proceed by induction: pick an element $a$, and two incompatible elements smaller than $a$, call them $a_0$ and $b_0$. Now $a'$ has two incompatible elements below it, so we can choose one to be $a_1$ and $b_1$. Proceed by induction splitting $b_n$ to $a_{n+1}$ and $b_{n+1}$. Then $\{a_n\mid n\in\Bbb N\}$ is the antichain you seek.

Choice is necessary, since without choice it is consistent there are counterexamples. For example, if $S$ is a set which is a countable union of pairs that no infinite set of pairs admits a choice function, then the tree of choice functions from finitely many pairs will satisfy this.

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    I think your assumption only makes sense for weak antichains, defined as pairwise incomparable. OP asked about strong antichains, defined as pairwise incompatible. – 2018-02-23
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    Nope, sorry. This is about antichains in the sense asked by the OP. At each step we are taking ***incompatible*** elements to form the antichain. But it's good that you pay attention before voting. – 2018-02-23
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    Your assumption that you only have to worry about elements where the order below them is linear is invalid. You need to take care of elements that are above a pairwise-compatible set, not pairwise-comparable. – 2018-02-23
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    Good reading comprehension skills. Read that single line again. – 2018-02-23
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    I'm confused about what you mean: you said "elements that below them the order is linear". That's pairwise-comparable, not pairwise-compatible. – 2018-02-23
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    Confused? No, you've made up your mind at least 8 minutes ago. Have a nice day. – 2018-02-23
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This is shown to be true for $\mathbb P=\tau\setminus\{\emptyset\}$ in Lemma 2.10 of [0]. It's likely known for general $\mathbb P$, but I can't find a citation, so I'll roll my own proof here. Assume $\mathbb P$ has (strong) antichains of size $n$ for all $n<\omega$. For compatible $p,q\in\mathbb P$, write $p\not\perp q$ and let $p\wedge q$ satisfy $p\wedge q\leq p,q$. If $p,q$ are incompatible, write $p\perp q$. Let $p^\downarrow=\{q\in\mathbb P:q\leq p\}$.

Say $p\in\mathbb P$ is bad if there exists $r_p\leq p$ such that $r_p^\downarrow$ is pairwise compatible. Let $\mathbb P_{bad}$ collect all bad points in $\mathbb P$, and say $p\sim q$ for $p,q\in\mathbb P_{bad}$ if $r_p\not\perp r_q$. This is obviously symmetric and reflexive, and if we assume $p\sim q,q\sim t$, then let $s_p=r_p\wedge r_q$ and $s_r=r_q\wedge r_t$. Since $r_q^\downarrow\in\mathbb P_{bad}$, $s_p\not\perp s_r$, so $r_p\not\perp r_t$ and thus $p\sim t$. Thus $\sim$ is an equivalence relation.

If $\mathbb P_{bad}/\sim$ is infinite, we may choose $p_i\in\mathbb P_{bad}$ such that $p_i\not\sim p_j$ for $i

Otherwise $|\mathbb P_{bad}/\sim|=n<\omega$, and choose an antichain $\{p_i:i\leq n\}$ in $\mathbb P$. If $\{p_i:i

So we've found $b_0\in\mathbb P\setminus\mathbb P_{bad}$. Given $b_n\in\mathbb P\setminus\mathbb P_{bad}$, we may choose $a_n,b_{n+1}\leq b_n$ such that $a_n\perp b_{n+1}$. Thus by construction, $a_n\perp a_{m+1}$ for all $n\leq m<\omega$. Therefore $\{a_n:n<\omega\}$ is an antichain.

[0]: W. W. Comfort and S. Negrepontis. Chain Conditions in Topology. Cambridge Tracts in Mathematics. Cambridge University Press, 1982.