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Let $P_o$ be $\pm$ odd primes and define a basis for open sets as $D(p) = \{ \pm(p-x) : x \geq p, x \in P_o\}$.

Proof: have $D(p \cup q) = D(p) \cap D(q)$ since $\pm (p-x) \in D(p \cup q) \iff x \geq p $ and $ x \geq q$ and $x \in P_o \iff x - p, p- x \in \{ \pm (p-x) : x \geq p\} \cap \{ \pm (p-x) : x\geq q\} = D(p) \cap D(q)$.

Therefore $D(\text{infinite set}) = \varnothing$ and so is $\bigcup\limits_{p \in P_o} D(p) = 2 \Bbb{Z}$?

$M = (\subset \Bbb{Z}, \cup, \cdot)$ forms a $\Bbb{Z}$-semimodule with elementwise multiplication by $\Bbb{Z}$, and $D : M \to N=(\subset\Bbb{Z}, \cap, \cdot)$ (where $D$ is defined more generally) is a semimodule hom.

Let $D$ be the more general one and let $D(x) \cap P_O$ be the odd prime one.

$D(\bigcup\limits_{x} ax) = a\bigcap\limits_{x} D(x)$

Daniel Donnelly
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1 Answers1

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No.

The inclusion $2 \mathbb Z \subseteq \bigcup_{p \in P_0} D(p)$ is equivalent to Schinzel's conjecture, which has not been proven yet. See also this question.

Fredrik Meyer
  • 20,228