Take a countably infinite set $S$, say $\mathbb N$. Is it possible for there to be an antichain in $\mathcal P(S)$ (with the inclusion ordering) of continuum cardinality?
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7$\begingroup$ It seems the question is already answered here. Also, a question like this is more appropriate for Mathematics Stack Exchange $\endgroup$– Saúl RMCommented May 25 at 1:48
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$\begingroup$ Here is an essay I had written about various features to be found in the lattice of sets of natural numbers: infinitelymore.xyz/p/lattice-of-sets-of-natural-numbers. $\endgroup$– Joel David HamkinsCommented May 25 at 23:59
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$\begingroup$ It's slightly incongruous to say that a question belongs somewhere else, and then to answer it as though it belongs here. $\endgroup$– Todd TrimbleCommented May 26 at 0:46
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$\begingroup$ Problem B-4 of the 1989 Putnam asked, "Can a countably infinite set have an uncountable collection of non-empty subsets such that the intersection of any two of them is finite?" (I suspect they copied this problem from Donald Newman's book, A Problem Seminar, since Problem A-4 from that year also appears in the same book.) I still remember Greg Landweber explaining to us afterward, "Take any real number, say $\pi$, and from it construct the set $\{3, 31, 314, 3141, 31415, \ldots\}$." By the way, another nice puzzle is, can there be a chain in $\mathcal{P}(S)$ of continuum cardinality? $\endgroup$– Timothy ChowCommented May 26 at 3:23
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2$\begingroup$ @SaúlRM Yes indeed, it's generally better to give an answer than just to answer in the comments, if the question belongs on the site. (I'd say this question is borderline, in the sense that it would have been widely considered acceptable for MO years ago, but not as much anymore.) Despite the incongruity, I wouldn't say you did anything wrong. What I would consider wrong is to both vote to close and to answer, but you didn't do that. Hope this all makes sense! $\endgroup$– Todd TrimbleCommented May 26 at 17:51
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Yes. Consider the countable set $\mathbb{Q}^2$ and for each $x\in\mathbb{R}$ the subset $\{(a,b)\in\mathbb{Q}^2;a\leq x,b\leq-x\}$.
Edit: See some easier examples in the comment below by bof. There are more examples of uncountable almost disjoint families (that is, uncountable antichains where the intersection between any two elements in the antichain is finite) in this MSE question.
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6$\begingroup$ Or the sets $\mathbb Q\cap(x,x+1)$ for $x\in\mathbb R$. Or the sets $X\subset\mathbb Z$ such that $|X\cap\{2n,2n+1\}|=1$ for all $n\in\mathbb Z $\endgroup$– bofCommented May 25 at 2:42
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$\begingroup$ Nice ones. The answer I link two above also gives an example where the intersection between any two elements of the antichain is finite $\endgroup$– Saúl RMCommented May 25 at 10:09