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# proof of alternative characterization of filter

First, suppose that $\mathbf{F}$ is a filter. We shall show that, for any two elements $A$ and $B$ of $\mathbf{F}$, it is the case that $A\cap B\in\mathbf{F}$ if and only if $A\in\mathbf{F}$ and $B\in\mathbf{F}$.

By the definition of filter, if $A\in\mathbf{F}$ and $B\in\mathbf{F}$ then $A\cap B\in\mathbf{F}$. Since $A\supseteq A\cap B$ and $\mathbf{F}$ is a filter, $A\cap B\in\mathbf{F}$ implies $A\in\mathbf{F}$. Likewise, $A\cap B\in\mathbf{F}$ implies $B\in\mathbf{F}$.

Next, we shall show that any proper subset $\mathbf{F}$ of the power set of $X$ such that $A\cap B\in\mathbf{F}$ if and only if $A\in\mathbf{F}$ and $B\in\mathbf{F}$ is a filter.

If the empty set were to belong to $\mathbf{F}$ then for any $A\subset X$, we would have $A\cap\emptyset=\emptyset\in\mathbf{F}$. This would imply that every subset of $X$ belongs to $\mathbf{F}$, contrary to our hypothesis that $\mathbf{F}$ is a proper subset of the power set of $X$.

If $A\subseteq B\subseteq X$ and $A\in\mathbf{F}$, then $A\cap B=A\in\mathbf{F}$. By our hypothesis, $B\in\mathbf{F}$.

The third defining property of a filter — If $A\in\mathbf{F}$ and $B\in\mathbf{F}$ then $A\cap B\in\mathbf{F}$ — is part of our hypothesis.

## Mathematics Subject Classification

03E99*no label found*54A99

*no label found*

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