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YCor
YCor
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Fixed acording to Benjamin's comment.
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André Caldas
André Caldas
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Take the integers $\mathbb{Z}$ and the addition \begin{align*} +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z} \\ (a,b) &\mapsto a+b. \end{align*} Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to \begin{equation*} +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z} \end{equation*} and then to \begin{equation*} +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}. \end{equation*} The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.

Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$. For any $p \in \mathbb{Z}^*$, we have the right ideal $I_p := p + \beta\mathbb{Z}$.

Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that \begin{equation*} p \in A \text{ and } q \in B \Rightarrow I_p \cap I_q = \emptyset? \end{equation*}

In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece?

Take the integers $\mathbb{Z}$ and the addition \begin{align*} +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z} \\ (a,b) &\mapsto a+b. \end{align*} Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to \begin{equation*} +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z} \end{equation*} and then to \begin{equation*} +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}. \end{equation*} The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.

Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$. For any $p \in \mathbb{Z}^*$, we have the ideal $I_p := p + \beta\mathbb{Z}$.

Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that \begin{equation*} p \in A \text{ and } q \in B \Rightarrow I_p \cap I_q = \emptyset? \end{equation*}

In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece?

Take the integers $\mathbb{Z}$ and the addition \begin{align*} +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z} \\ (a,b) &\mapsto a+b. \end{align*} Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to \begin{equation*} +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z} \end{equation*} and then to \begin{equation*} +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}. \end{equation*} The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.

Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$. For any $p \in \mathbb{Z}^*$, we have the right ideal $I_p := p + \beta\mathbb{Z}$.

Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that \begin{equation*} p \in A \text{ and } q \in B \Rightarrow I_p \cap I_q = \emptyset? \end{equation*}

In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece?

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André Caldas
André Caldas
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Partitioning $\beta \mathbb{Z} \setminus \mathbb{Z}$

Take the integers $\mathbb{Z}$ and the addition \begin{align*} +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z} \\ (a,b) &\mapsto a+b. \end{align*} Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to \begin{equation*} +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z} \end{equation*} and then to \begin{equation*} +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}. \end{equation*} The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.

Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$. For any $p \in \mathbb{Z}^*$, we have the ideal $I_p := p + \beta\mathbb{Z}$.

Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that \begin{equation*} p \in A \text{ and } q \in B \Rightarrow I_p \cap I_q = \emptyset? \end{equation*}

In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece?