2 reaction to Joel's comment

To get the actual answer out of the way: the usual definition of ideal implies that any ideal contains the empty set -- an ideal $I$ (on a set $X$ / the power set of $X$) is non-empty, closed under taking subsets and under taking finite unions (and of course $I\subseteq P(X)$). The first two should convince you that it contains the empty set.

The notion of filter is dual -- non-empty, closed under taking finite intersections and supersets. Ideals correspond to filters by mapping each element $A\in I$ to its complement $X\setminus A$. Maximal ideals correspond to maximal filters.

Just in case this question was only due to a confusion of ideals and filters, let me add:

A proper ideal by definition does not contain the 'full' set $X$ (e.g. in your example $X = { 1,2,3 }$). Similarly, a proper filter does not contain the empty set by definition. The 'improper' cases of these definitions coincide -- both the improper filter and the improper ideal are just the full power set (as is clear from being closed under subsets and supersets respectively).

Usuallythe interest lies in the , filter/ideal means propercases filter/ideal, but sometimes for notational or technical convenience, it is sometimes seems nice to allow the improper case -- for example in the Stone-Cech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed non-empty subsets and the improper filter to the empty set. But I think the general preference (as Joel David Hamkins pointed out in the comments) is not to do this since no convenience outweighs the confusion caused by the improper case.

In your example and (as mentioned by Robin Chapman) for any finite set $X$ , the maximal (proper) filters (or ultrafilters) are the principal filters, i.e. those of the form $\dot{x} = { A \subseteq X:\ x \in A }$ for some $x\in X$. To see this just partition $X$ into singletons -- a finite partition by assumption on $X$ -- every maximal filter contains exactly one part of the partition. The maximal ideals are again the dual.

1

To get the actual answer out of the way: the usual definition of ideal implies that any ideal contains the empty set -- an ideal $I$ (on a set $X$ / the power set of $X$) is non-empty, closed under taking subsets and under taking finite unions (and of course $I\subseteq P(X)$). The first two should convince you that it contains the empty set.

The notion of filter is dual -- non-empty, closed under taking finite intersections and supersets. Ideals correspond to filters by mapping each element $A\in I$ to its complement $X\setminus A$. Maximal ideals correspond to maximal filters.

Just in case this question was only due to a confusion of ideals and filters, let me add:

A proper ideal by definition does not contain the 'full' set $X$ (e.g. in your example $X = { 1,2,3 }$). Similarly, a proper filter does not contain the empty set by definition. The 'improper' cases of these definitions coincide -- both the improper filter and the improper ideal are just the full power set (as is clear from being closed under subsets and supersets respectively).

Usually the interest lies in the proper cases, but sometimes for notational or technical convenience, it is nice to allow the improper case -- for example in the Stone-Cech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed non-empty subsets and the improper filter to the empty set.

In your example and (as mentioned by Robin Chapman) for any finite set $X$ , the maximal (proper) filters (or ultrafilters) are the principal filters, i.e. those of the form $\dot{x} = { A \subseteq X:\ x \in A }$ for some $x\in X$. To see this just partition $X$ into singletons -- a finite partition by assumption on $X$ -- every maximal filter contains exactly one part of the partition. The maximal ideals are again the dual.