Kuratowski's Free Set Theorem

Theorem
Let $n$ be a positive integer and let $ X$ be an infinite set. Then the cardinality of $X$ is greater than or equal to $\aleph_n$ if and only if for every mapping $f$ from $[X]^n$ (the family of $n$-element subsets of $X$) into $[X]^{<\aleph_0}$ (the family of finite subsets of $X$) there exists an  $(n + 1)$-element free subset of $X$ with respect to $f$, i.e., a subset $Y$ of $X$ with $n+1$ elements  such that $y\notin f(Y\setminus\{ y\})$ for all $y\in Y$.

Proof, part 1
We show that the conclusion does not hold if $X$ has cardinality $<\aleph_n$.

For each $n>0$ we will find a function $f_n:[\omega_{n-1}]^n \to [\omega_{n-1}]^{<\aleph_0}$ which has no free set of size $n+1$. The restriction of $f$ to any subset of $\omega_{n-1}$ will then witness that this subset does not satisfy the conclusion of the theorem.

For $n=1$, let $f_1:\omega \to [\omega]^{<\aleph_0}$ be defined by $f_1(n) = \{0,\ldots, n-1\}$. Then for every $\{k,n\}\in [\omega]^2$ we have $k<n$ or $n<k$, so  $K\in f(n)$ or $n\in f(k)$,  hence the set $\{k,n\}$ is not free.

We continue by induction. Assume that for every set $A$ of size $<\aleph_n$ there is a function $f_n^A:[A]^n\to [A]^{<\aleph_0}$ which has no free set of size $n+1$. We define a function $f_{n+1}:[\omega_n]^{n+1}\to [\omega_n]^{<\aleph_0}$ as follows: (Note that $z$ is an ordinal $< \omega_n$, hence a set of size $<\aleph_n$, so $f_n^y$ is a well-defined function on $[y]^n$; the set $Y$ is in the domain of this function.)
 * Every set if $[\omega_n]^{n+1}$ can be written as $Y\cup \{z\}$ with $Y \in [\omega_n]^n$, $z\in \omega_n$, $z>\max Y$.
 * We let $f_{n+1}(Y\cup \{z\}) := f_n^z(Y)$.

To conclude the proof, we claim that no set $B:=\{y_0 < y_1<\cdots < y_n < z\}$ can be free for $f_{n+1}$. By definition, the function $f_n^z$ has no free set of size $n+1$. So in particular, the set $\{y_0,\ldots, y_n\}$ is not free for $f_n^z$, say $y_i \in f_n^z(y_0,\ldots, y_{i-1},y_{i+1},\ldots, y_n)$. But then also $y_i \in f_{n+1}(B\setminus y_i)$, hence $B$ is not free.

Proof, part 2
We show that the conclusion holds if $|X|\ge\aleph_n$. In this case, $X$ will have a subset $Y$ of cardinality $\aleph_n$; it is enough to find a free subset for every function $f$ on $[Y]^{n+1}$.

For $n=1$, let $f:\omega_1\to [\omega_1]^{<\aleph_0}$ be any function. The set $$ S:= \{\delta < \omega_1 \mid \delta\text{ limit}, \forall \alpha < \delta: f(\alpha) \subseteq \delta\}$$ is certainly an unbounded set (in fact it is a closed unbounded subset of $\omega_1$): For every $\alpha_0< \omega_1$ the supremum of the $\omega$-sequence $\alpha_0 < f^*(\alpha_0) < f^*(f^*(\alpha_0)) < \cdots $ will be in $S$.

So let $\delta\in S$ be infinite, and let $\alpha<\delta $ be such that $\alpha\notin f(\delta)$. Then also $\delta \notin f(\alpha)$, so $\{\alpha,\delta\}$ is free for $f$.

Again we continue by induction. Let $n\ge 2$, and let $f:[\omega_n]^n\to [\omega_n]^{<\aleph_0}$ be any function. As before, the set $S:= \{ \delta < \omega_1\mid \delta\text{ limit}, \forall \alpha_1<\cdots<\alpha_n< \delta: f(\alpha_1,\ldots, \alpha_n) \subseteq \delta\}$ is unbounded. Let $\delta\in S$, $\delta \ge \omega_{n-1}$.

Now the function $f':[\delta]^{n-1}\to [\delta]^{<\aleph_0}$, defined by $f'(x_1,\ldots, x_{n-1}) := f(x_1,\ldots, x_{n-1},\delta)$ is an $n-1$-ary function on a set of size $\aleph_{n-1}$, so by induction hypothesis there is a set $A=\{\alpha_1< \cdots < \alpha_n\}$ which is free with respect to $f'$.

We claim that $A\cup \{\delta\} = \{\alpha_1< \cdots < \alpha_n<\delta\}$ is free for $f$. By definition of $\delta$ we have $\delta\notin f(\alpha_1,\ldots, \alpha_n)$, and the freeness of $A$ with respect to $f'$ ensures $\alpha_i \notin f'(A\setminus \{\alpha_i\} ) = f(A \setminus\{\alpha_i\}\cup \{\delta\})$.