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There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter.

But there's another, shorter way to state the definition:

Let $X$ be a set. An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions $$ X = X_1 \amalg \cdots \amalg X_n $$ of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$.

I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down. Can anyone help?

Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3. Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$. (The same thing works with 4, or 5, etc., though not with 2.) I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too.

Edit To be clear, when I use the word "partition" I don't mean to imply that the sets $X_i$ are nonempty. I just mean a finite family of pairwise disjoint sets $X_i$ whose union is $X$. They can be empty.

There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter.

But there's another, shorter way to state the definition:

Let $X$ be a set. An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions $$ X = X_1 \amalg \cdots \amalg X_n $$ of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$.

I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down. Can anyone help?

Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3. Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$. (The same thing works with 4, or 5, etc., though not with 2.) I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too.

Edit To be clear, when I use the word "partition" I don't mean to imply that the sets $X_i$ are nonempty. I just mean a family of pairwise disjoint sets $X_i$ whose union is $X$. They can be empty.

There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter.

But there's another, shorter way to state the definition:

Let $X$ be a set. An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions $$ X = X_1 \amalg \cdots \amalg X_n $$ of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$.

I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down. Can anyone help?

Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3. Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$. (The same thing works with 4, or 5, etc., though not with 2.) I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too.

There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter.

But there's another, shorter way to state the definition:

Let $X$ be a set. An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions $$ X = X_1 \amalg \cdots \amalg X_n $$ of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$.

I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down. Can anyone help?

Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3. Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$. (The same thing works with 4, or 5, etc., though not with 2.) I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too.

Edit To be clear, when I use the word "partition" I don't mean to imply that the sets $X_i$ are nonempty. I just mean a finite family of pairwise disjoint sets $X_i$ whose union is $X$. They can be empty.