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Let $n$ be a positive natural number. Prove using Zorn's lemma that there is a set A of points in the plane that satisfies: 1. Any line in the plane does not contain $n+1$ points of A. 2. For every point in the plane that isn't in A, there is at least one line in the plane that contains her and $n$ more points of A. Denoting the set of all set that satisfies statement number 2 by B and order it via the order $\subset$. Let $S$ be a chain in $(B,\subset)$. I've tried showing that $\cup S\in B$ but I haven't succeeded. I would like a hint.

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Hint: Use Zorn's Lemma to find a set that is maximal with respect to property (1). The basic idea here is that if $\bigcup S$ fails to satisfy (1), then you should be able to find, using the fact that $S$ is a chain with respect to $\subseteq$, a set in $S$ which also fails to satisfy (1). After this, show that such a set must satisfy property (2).

(There is obviously a set maximal with respect to property (2): the set of all points in the plane!)

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So that's how we're gonna do it? I finally get some sleep and you snag this away from me? I demand pickled taxation! :-P –  Asaf Karagila Feb 6 at 16:02
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@Asaf: It had been open for, like, eight minutes! What was I supposed to do? Not answer it? –  Arthur Fischer Feb 6 at 16:23

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