Added the principal logarithm.
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@@ -47,6 +47,25 @@
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(2) $\Rightarrow$ (1): Let $E, F \in \tau$, then $E \cup F \in \sigma$. Since $\tau$ is fundamental, there exists $G \in \tau$ such that $E \cup F \subset G$.
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\end{proof}
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\begin{definition}[Product Ideal]
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\label{definition:product-ideal}
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Let $X, Y$ be sets, $\sigma \subset 2^X$ and $\tau \subset 2^Y$ be ideals, and
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\[
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\beta = \bracs{A \times B|A \in \sigma, B \in \tau}
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\]
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then there exists a unique ideal $\sigma \times \tau$ such that $\beta$ is fundamental with respect to $\sigma$. The ideal $\sigma \otimes \tau$ is the \textbf{product} of $\sigma$ and $\tau$.
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\end{definition}
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\begin{proof}
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For each $A_1, A_2 \in \sigma$ and $B_1, B_2 \in \tau$,
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\[
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(A_1 \times B_1) \cup (A_2 \times B_2) \subset (A_1 \cup A_2) \times (B_1 \cup B_2)
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\]
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By \autoref{proposition:set-ideal-fundamental-criterion}, there exists an ideal $\sigma \otimes \tau$ such that $\beta$ is fundamental with respect to $\sigma \otimes \tau$.
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\end{proof}
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