Proposition 7.1.4.label Let $X$ be a set and $\tau \subset 2^{X}$, then the following are equivalent:

  1. (1)

    For any $E, F \in \tau$, there exists $G \in \tau$ such that $E \cup F \subset G$.

  2. (2)

    There exists an ideal $\sigma \subset 2^{X}$ such that $\tau$ is fundamental with respect to $\sigma$.

If the above holds, then the ideal $\sigma$ in (2) is the ideal generated by $\tau$.

Proof. (1) $\Rightarrow$ (2): Let

\[\sigma = \bracs{F \subset X| \exists E \in \tau: F \subset E}\]

then $\sigma$ satisfies (I1) by definition. For any $E, F \in \sigma$, there exists $E_{0}, F_{0} \in \tau$ such that $E \subset E_{0}$ and $F \subset F_{0}$. By assumption, there exists $G \in \tau$ such that

\[E \cup F \subset E_{0} \cup F_{0} \subset G\]

so $\sigma$ satisfies (I2), and is an ideal.

(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 $E \cup F \subset G$.$\square$

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