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Suppose $S$ is a k-ary relation on $\mathbb{N}$. Given $1 \leq i \leq k$, define $\pi_i(S)$ in the following way: $$\left \{s_i | (s_1, s_2, \cdots s_i \cdots s_k) \in S \right \}$$.

We call $S$ nice if: for every component $i$, either the projection $\pi_i(S)$ or its complement is finite.

I want to show that if $S$ is nice, then so is $\mathbb{N}^k \setminus S$.

I came across this problem while trying to solve another problem in logic.

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Let $k = 2$, $S = \{ (x, y) : (2 \nmid x) \text{ or } y = 0\}$, and $S' = \mathbb{N}^2 \setminus S = \{ (x, y) : (2 \mid x) \text{ and } y \ne 0\}$. Then $\mathbb{N} \times \{0\} \subset S$, $\{1\} \times \mathbb{N} \subset S$, so $\pi_1(S) = \pi_2(S) = \mathbb{N}$, so $S$ is nice. But $\pi_1(S') = \{ x \in \mathbb{N} : (2 \mid x) \}$, and neither this set nor its complement is finite, therefore $S'$ is not nice; so the conjecture is false.

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  • $\begingroup$ This was not the definition of niceness. We require the projections to be finite or co-finite, not the relation itself $\endgroup$ – Agnishom Chattopadhyay Dec 2 '17 at 2:29
  • $\begingroup$ Sorry, my bad. Upvoted. $\endgroup$ – Agnishom Chattopadhyay Dec 2 '17 at 4:06

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