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KP Hart
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Let $2^{<\omega}$ be the binary tree and assign to each branch $x$ the family $F_x$ of finite sets that intersect it. If $x_1$, $x_2$, $\ldots$ $x_k$ is a finite set of (distinct) branches then there is a level, $n$ say, where they all differ. On each level above $n$ you can find finite sets that satisfy any Boolean combination of the $F_{x_i}$ you like.

Let $2^{<\omega}$ be the binary tree and assign to each branch $x$ the family $F_x$ of finite sets that intersect it. If $x_1$, $x_2$, $\ldots$ $x_k$ is a finite set of (distinct) branches then there is a level, $n$ say, where they all differ. On each level you can find finite sets that satisfy any Boolean combination of the $F_{x_i}$ you like.

Let $2^{<\omega}$ be the binary tree and assign to each branch $x$ the family $F_x$ of finite sets that intersect it. If $x_1$, $x_2$, $\ldots$ $x_k$ is a finite set of (distinct) branches then there is a level, $n$ say, where they all differ. On each level above $n$ you can find finite sets that satisfy any Boolean combination of the $F_{x_i}$ you like.

Source Link
KP Hart
  • 11.4k
  • 38
  • 48

Let $2^{<\omega}$ be the binary tree and assign to each branch $x$ the family $F_x$ of finite sets that intersect it. If $x_1$, $x_2$, $\ldots$ $x_k$ is a finite set of (distinct) branches then there is a level, $n$ say, where they all differ. On each level you can find finite sets that satisfy any Boolean combination of the $F_{x_i}$ you like.