CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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The union of the two principal ultrafilters on a two-point set is not a filter

Statement refuted

The union of any two filters on a set is again a filter.

On $X = {0, 1}$ , let $U_{0}$ and $U_{1}$ be the principal ultrafilters at $0$ and $1$ . Their union is not a filter.

Facts & Assumptions

Given: The set $X = {0, 1}$ and the principal ultrafilters $U_{0} = {A \subseteq X : 0 \in A}$ and $U_{1} = {A \subseteq X : 1 \in A}$ .

[L1]

For every $x \in X$ , the subsets of $X$ containing $x$ form the principal ultrafilter $U_{x}$ (The subsets of $X$ containing a fixed point $x$ form the principal ultrafilter at $x$ ).

[F1]

A filter is closed under pairwise intersection and omits $\emptyset$ (Filter on a set).

[L2]

The union of a nonempty inclusion-chain of filters is a filter; comparability is used to place any two members in one filter before intersecting them (The union of a nonempty chain of filters is a filter).

Counterexample

technique · direct

1.1

By [L1], $U_{0}$ and $U_{1}$ are filters on $X$ .

givenL1

1.2

The singleton ${0}$ belongs to $U_{0}$ and the singleton ${1}$ belongs to $U_{1}$ , so both belong to $U_{0} \cup U_{1}$ .

given

1.3

The two filters are not comparable: ${0} \in U_{0} ∖ U_{1}$ and ${1} \in U_{1} ∖ U_{0}$ . Hence [L2] does not apply, and the example shows why its chain hypothesis is essential.

givenL2

2.1

Neither principal ultrafilter contains $\emptyset$ , so $\emptyset \notin U_{0} \cup U_{1}$ .

step 1.1F1

3.1

If $U_{0} \cup U_{1}$ were a filter, intersection closure applied to ${0}$ and ${1}$ would put $\emptyset = {0} \cap {1}$ in the union, contradicting step 2.1. Thus the union is not a filter.

step 1.2step 2.1F1

4.1

Therefore an arbitrary union of two filters, even two principal ultrafilters, need not be a filter.

step 1.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Filter (set theory) (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)