울트라필터 (set 이론)

In the mathematical field of set theory, an ultrafilter is a maximal proper filter: it is a filter ${\displaystyle U}$ on a given non-empty set ${\displaystyle X}$ which is a certain type of non-empty family of subsets of ${\displaystyle X,}$ that is not equal to the power set ${\displaystyle \wp (X)}$ of $$${\displaystyle X}$ ${\displaystyle X}($이러한 필터를 적절한 필터라고 함) 및 $X{\displaystyle }$ ${\displaystyle$ X$}$에 적절한 하위 집합으로 포함하는$$ 다른 적절한 필터가 없다는 점에서 "최대"이기도 하다.다르게 말하면 적절한 필터 $U{\displaystyle }$ ${\displaystyle$ U$}$이(가) $U{\displaystyle }$ ${\displaystyle U}$ 그 자체인$$ 적절한 필터(필요하게)가 정확히 하나만 있으면 울트라필터라고 한다$$.

$좀{\displaystyle }$ 더 공식적으로 X ${\displaystyle$ X$}$의 U ${\displaystyle$ U$}$은(는) X {\$displaystyle$ X$}$의 최대 필터이기도 $하며$, 이는$$ U ${\displaystyle U}$을(를) 적절한 하위$$ $집합$으로 $포함$하는 X ${\displaystyty$ X}에 $적절한$필터가 없다는 것을 의미한다.세트의 Ultrafilters on sets는 부분적으로 정렬된 세트의 중요한 특수한 Ultrafilter 사례로, 여기서 부분 순서는 전원 세트${\displaystyle \mathrm{}}$℘${\displaystyle }$( ${\displaystyle X}$) ${\displaystyle \wp (X)}$로 구성되며$$ 부분 순서는 부분 집합 포함 ${\displaystyle \subseteq .}$ ${\displaystyle \,\subseteq .}$

울트라필터는 세트 이론, 모델 이론, 토폴로지에 많은 응용 프로그램을 가지고 있다.^{[1]: 186}

정의들

임의 집합 $X{\displaystyle }$, ${\displaystyle X,}$이(가) X ${\displaystyle X}$에 있는 울트라필터는 $X{\displaystyle }$ ${\displaystyle$ X$}$의 $부분$ 집합 중$$ 비어 있지 않은 패밀리 $U{\displaystyle }$ ${\displaystyle U}$이며, 다음과 같은 경우$$:

1. 적절 또는 비감소:빈 집합은 $U{\displaystyle .}$ ${\displaystyle$ U$.}$의 요소가 아님
2. Upward closed in ${\displaystyle X}$: If ${\displaystyle A\in U}$ and if ${\displaystyle B\subseteq X}$ is any superset of ${\displaystyle A}$ (that is, if ${\displaystyle A\subseteq B\subseteq X}$) then ${\displaystyle B\in U.}$
3. π-시스템:$A{\displaystyle }$ ${\displaystyle A}$ $및$ B ${\displaystyle B}$이(가) $U{\displaystyle }$ ${\displaystyle U}$의 요소인$$ 경우 $이들$의 교차로 A ${\displaystyle \cap }$ ${\displaystyle B.}$ ${\displaystyle A\cap B$도 마찬가지다$.$$}$
4. $A{\displaystyle }$ ${\displaystyle \subseteq }$ ${\displaystyle X}$ ${\displaystyle A\subseteq$ X$}$인^{[note 1]} $경우{\displaystyle }$ A ${\displaystyle A}$ 또는$$ 그 상대적 보완물 $X{\displaystyle }$ ${\displaystyle \setminus }$ ${\displaystyle A}$ ${\displaystyle X\setminus$ A$}$은(는) $U{\displaystyle }$ .${\displaysty$ U$.}$의 요소임.

속성 (1), (2) 및 (3)은 $X{\displaystyle .}$ ${\displaystyle X.}$에 대한 필터의 정의 속성이다. 일부 저자는$$ "필터"의 정의에 비지연성(위의 속성 (1))을 포함하지 않는다.그러나, "초광필터"(그리고 "프리필터"와 "필터 서브베이스"의 정의도 항상 정의 조건으로 비지연성을 포함한다.이 글은 필터가 강조를 위해 "속성"으로 설명될 수 있지만 모든 필터가 적절해야 한다고 요구한다.

일체형이 아니ultrafilter 필터 F{F\displaystyle}로, 사람 말할 것이다 m(A)=1{\displaystyle m(A)=1}A∈ F{A\in F\displaystyle}과 m(A)=0{\displaystyle m(A)=0}만약 X∖ ∈ F,{\displaystyle X\setminus A\in F.} 떠나m{m\displaystyle}한정되지 않은 다른 곳.[표창 필요한][해명 필요한]

필터 서브베이스는 유한 교차로 특성을 갖는 비빈 집합 집합(즉, 모든 유한 교차로들이 비빈다)이다.마찬가지로 필터 하위 베이스는 일부(속성) 필터에 포함된 비어 있지 않은 세트 제품군이다.주어진 필터 서브베이스를 포함하는 가장 작은 필터$({\displaystyle \subseteq$ $\}\}$에 상대적인 필터는 필터 서브베이스에 의해 생성된다고 한다.

집합 $P{\displaystyle }$ ${\displaystyle P}$ $집합$의 X ${\displaystyle X}$에서 위쪽 닫힘이 설정된$$ 경우

${\displaystyle P^{\uparrow X}=\왼쪽\{S:A\subseteq S\subseteq X{\text{일부 }}A\in P\right\}.}$

A prefilter or filter base is a non-empty and proper (i.e. ${\displaystyle \varnothing \not \in P}$) family of sets ${\displaystyle P}$ that is downward directed, which means that if ${\displaystyle B,C\in P}$ then there exists some ${\displaystyle A\in P}$ such that ${\displaystyle A$$\subseteq B\cap C.}$ Equivalently, a prefilter is any family of sets ${\displaystyle P}$ whose upward closure ${\displaystyle P^{\uparrow X}}$ is a filter, in which case this filter is called the filter generated by ${\displaystyle P}$ and ${\displaystyle P}$ is said to be a filter base for ${\displaystyle P^{\uparrow X}.}$$P^{\displaystyle P^{\uparrow X}.$$}$

세트 $P{\displaystyle }$ ${\displaystyle P}$ 계열의$$^[2] $X{\displaystyle }$ ${\displaystyle X}$에 있는 이중은$$ $세트{\displaystyle }$ X${\displaystyle }$: ${\displaystyle P}$ ${\displaystyle :=}$ { ${\displaystyle Xb}$B : ${\displaystyle B}$ ${\displaystyle \in }$ ${\displaystyle \}}$ } } } }${\displaystyle }$} . ${\displaystyle$ X$\setminus$ P$:=\좌측\{X\setminus B:::$$B\in P\right\}$$}$

울트라 프리필터에 대한 일반화

$X{\displaystyle }$ ${\displaystyle X}$의 하위 집합 중$$ U \${\displaystyle U\neq \varnot }$ 계열 $U{\displaystyle }$ ∉ ${\displaystyle \notin }$ ${\displaystyle U}$ ${\displaysty \varnot \in$ U$}$과(와)의$$ 어느 하나라도 충족되면 Ultra라고 부른다$$.^[2][3]

1. For every set ${\displaystyle S\subseteq X}$ there exists some set ${\displaystyle B\in U}$ such that ${\displaystyle B\subseteq S}$ or ${\displaystyle B\subseteq X\setminus S}$ (or equivalently, such that ${\displaystyle B\cap S}$ equals ${\displaystyle B}$ or ${\displa$$ystyle \varnothing }$ $$.
2. For every set ${\displaystyle S\subseteq \bigcup _{B\in U}B}$ there exists some set ${\displaystyle B\in U}$ such that ${\displaystyle B\cap S}$ equals ${\displaystyle B}$ or ${\displaystyle \varnothing .}$
   • 여기서 $\bigcup {\displaystyle \underset{}{}}$ ${\displaystyle \underset{}{B}}$ ${\displaystyle \underset{}{\in }}$ ${\displaystyle \underset{}{}}$ B ${\displaystyle \bigcup _{B\in U}B}$은(는) $U{\displaystyle }$. {\$displaystyle U.}$에 있는 모든 세트의 조합으로 정의된다$$.
   • "${\displaystyle }$ ${\displaystyle U}$의 이러한 특성은 $X{\displaystyle }$ , {\$displaystyle$ $X$$}$ 집합에 의존하지 않으므로 "ultra$$"라는 용어를 사용할 때 X ${\displaystyle$ X$}$ 집합에 $대한{\displaystyle }$ 언급은 선택 사항이다$$.
3. 모든 세트 $S{\displaystyle }$ ${\displaystyle$ S$}($$X{\displaystyle }$ ${\displaystyle X$의 부분 집합도 아님)에 대해 일부 세트 $B{\displaystyle }$ $∈$${\displaystyle }$U ${\$$displaystyle$ ${\displaystyle B}$$\cap$S$}이($$가$$)$ B ${\displaystystyle B}$ 또는$$ ${\displaystystyle \varnote.}$과(으)가 있다.
   • If ${\displaystyle U}$ satisfies this condition then so does every superset ${\displaystyle V\supseteq U.}$ In particular, a set ${\displaystyle V}$ is ultra if and only if ${\displaystyle \varnothing \not \in V}$ and ${\displaystyle V}$ contains as a subset some ultra family of sets.

울트라 필터 서브베이스는 반드시 프리필터다.^{[proof 1]}

Ultra Property는 이제 Ultra Filter와 Ultra Prefilter를 모두 정의하는 데 사용할 수 있다.

울트라 프리필터는^[2][3] 울트라 프리필터다.동등하게, 초고속의 필터 서브베이스다.

$X{\displaystyle }$ ${\displaystyle X}$의 울트라필터는^[2][3] $X{\displaystyle }$ ${\displaystyle X}$의 (속성) 필터로$$, 울트라다$$.마찬가지로, 이 필터는 $X{\displaystyle }$ ${\displaystyle X}$의 모든 필터로$$, 울트라 프리필터에 의해 생성된다.

대형 집합으로 해석

X{X\displaystyle}에 대한 적절하고 필터 F{F\displaystyle}의 요소로" 큰 세트(F를{\displaystyle F}상대적)"의 그리고 큰 세트의 X{X\displaystyle}의 보완을 함" 작은"sets[4]이상적인 X에서 자세한 내용은" 작은 세트"던 그대로 요소 ∖ F의 생각될 수 있는 것으로 볼 수 있 {)$디스플레이$ 스타일 $X\setminus F}$ $$.일반적으로 크지도 작지도 않은 $X$ ${\displaystyle X}$의 하위 집합이 있을 수 있으며, 크지도 작을 수도 있다.이중 이상은 크고 작은 세트가 없거나, equival${\displaystyle \varnothing}$이(가) 크지 않은 경우 필터(즉, 적절한)이다.^[4]필터는 $X{\displaystyle }$ ${\displaystyle X}$의 모든 하위 집합이 크거나$$ 작은 경우에만 울트라다.With this terminology, the defining properties of a filter can be restarted as: (1) any superset of a large set is large set, (2) the intersection of any two (or finitely many) large sets is large, (3) ${\displaystyle X}$ is a large set (i.e. ${\displaystyle F\neq \varnothing }$), (4) the empty set is not large.서로 다른 이중 이상은 "크게"라는 다른 개념을 준다.

Another way of looking at ultrafilters on a power set ${\displaystyle \wp (X)}$ is as follows: for a given ultrafilter ${\displaystyle U}$ define a function ${\displaystyle m}$ on ${\displaystyle \wp (X)}$ by setting ${\displaystyle m(A)=1}$ if ${\displaystyle A}$ is an element o그렇지 않으면$$ ${\displaystyle \mathrm{F}}$ ${\displaystyle U}$ 및 $m$(A )${\displaystyle }$= 0$${\$displaystyle$ m$(A$)=$0}$이러한 함수를 2값 형태론이라고 한다.그러면 $m{\displaystyle }$ ${\displaystyle m}$은(는) 미세하게 첨가되며$$, 따라서 $\wp {\displaystyle \mathrm{}}$( X${\displaystyle }$)${\displaystyle }$, ${\displaystyle \wp (X)$ 및 $X$ ${\displaystyle X}$ 요소의 모든 속성은 거의 모든 곳에서 참이거나 거의 모든 곳에서 거짓이다.그러나 $m{\displaystyle }$ ${\displaystyle m}$은(는) 대개 계산적으로 가법적으로 첨가되지 않으므로$$ 일반적인 의미에서 측정치를 정의하지 않는다.

최대 프리필터로서의 울트라 프리필터

"최대성"이라는 관점에서 울트라 프리필터를 특성화하기 위해서는 다음과 같은 관계가 필요하다.

만약 항의라도,{\displaystyle N,}세트 M{M\displaystyle}과 N의 두 가족으로 볼 때 가족 M{M\displaystyle}N보다,{\displaystyle N,}와 N{N\displaystyle}보다 M에 종속되finer은 coarser[5][6]고 있는{\displaystyle M,}, M⊢ M≤ N{M\leq N\displaystyle}또는 N을 쓴 것이다.r지금까지y ${\displaystyle X\in M,}$ there is some ${\displaystyle F\in N}$ such that ${\displaystyle F\subseteq C.}$ The families ${\displaystyle M}$ and ${\displaystyle N}$ are called equivalent if ${\displaystyle M\leq N}$ and ${\displaystyle N\leq M.}$ The families ${\displaystyle M}$$$이러한 세트 중 하나가 다른 세트보다 미세한 경우 ${\displaystyle M}$과(와) $N{\displaystyle }$ ${\displaystyle$ N$}$이(가) 비교 가능하다.^[5]

종속 관계, 즉 ${\displaystyle \ge }$, ${\$은(는) 사전 순서여서 위의 "등가" 정의는 동등성 관계를 형성한다.$M{\displaystyle }$ ${\displaystyle \subseteq }$ ${\displaystyle N}$ ${\displaystyle M\subseteq N}$인 경우, $M{\displaystyle }$n ${\displaystyle N}${\$displaystyle$M\$leq$ N$}$을(를) 사용하지만$$, 이 역은 일반적으로 유지되지 않는다.그러나 필터와 같이 $N{\displaystyle }$ ${\displaystyle N}$이(가) 위쪽으로 닫힌 경우$$ $M{\displaystyle }$${\displaystyle \le }$ ${\displaystyle N}$ ${\displaystyle M\leq$ N$}$인 경우$$, 모든$$ 프리필터는 생성되는 필터와 동일하다.이는 필터가 필터가 아닌 세트와 동등한 것이 가능하다는 것을 보여준다.

$두{\displaystyle }$ $집합$의$$M {\$displaystyle$M$$}과$$N {\$displaystyle$N}이(가) 동일하면$$ M ${\displaystyle M}$과 $N{\displaystyle }$ ${\displaystyle$ N$}$이(가) 모두$$ 울트라(resp. 프리필터, 필터 서브베이스)이거나 둘 다 울트라(resp.특히 필터 서브베이스가 프리필터가 아닌 경우, 필터 서브베이스가 생성하는 프리필터나 프리필터와 동일하지 않다.$M{\displaystyle }$ ${\displaystyle M}$과 $N{\displaystyle }$ ${\displaystyle N}$이(가) $모두{\displaystyle }$ X ${\displaystyle X}$의 필터인$$ 경우 $M{\displaystyle }$ ${\displaystyle M$$}$과 $N{\displaystyle }$ ${\$$displaystystyle N}$이${\displaystyle (}$가) 적절한 필터(resp)인$$ 경우에만 동일하다$.$울트라필터)는 세트 $M{\displaystyle }$ ${\displaystyle M}$과 동일하며, $M$ ${\displaystyle M}$은 반드시 프리필터(resp)이다$$.울트라 프리필터).다음과 같은 특성을 이용하여 프리필터(resp)를 정의할 수 있다.필터(resp)의 개념만을 사용하는 초프리필터.울트라필터) 및 후순위:

임의의 세트 패밀리는 (속성) 필터와 동등한 경우에만 프리필터다.

임의의 세트 패밀리는 만약 그것이 초필터와 같다면 그리고 그것만이 초필터에 해당한다.

$X{\displaystyle }$ ${\displaystyle X}$의 최대 프리필터는 다음과 같은 동등한 조건 중 하나를 충족하는$$ 프리필터 $U{\displaystyle }$ $){\displaystyle U\subseteq \wp(X)}$이다^[2][3].

1. ${\displaystyle U}$ ${\displaystyle U}$은(는) 울트라다.
2. ${\displaystyle U}$ is maximal on ${\displaystyle \operatorname {Prefilters} (X)}$ with respect to ${\displaystyle \,\leq ,}$ meaning that if ${\displaystyle P\in \operatorname {Prefilters} (X)}$ satisfies ${\displaystyle U\leq P}$ then ${\displaystyle$$P\leq U.}$^[3]
3. $U{\displaystyle .}$ ${\displaystyle U.}$에 적절하게 종속된 프리필터가 없음
4. $X{\displaystyle }$ ${\displaystyle X}$의 (property) $필터{\displaystyle }$ F {\$displaystyle U\leq$ $F$$}$을(를) ${\displaystyle \mathrm{만족}}$하면 $F$$$${\displaystyle \le }$ ${\displaystyle U.}$ ${\displaystyle F\leq$ U$.}$
5. $U{\displaystyle }$ ${\displaystyle U}$이$($가) 생성하는$$X {\$displaystyle$ X}의 필터는 울트라다$$.

화화화

빈 세트에는 울트라필터가 없으므로, $따라서{\displaystyle }$ X ${\displaystyle$ X$}$이(가) 비어 있지 않은 것으로 가정한다.

$X{\displaystyle }$ ${\displaystyle X}$의 필터 하위 베이스 $U{\displaystyle }$ ${\displaystyle U}$은(는) $다음{\displaystyle }$ 동등한 조건 중 하나가 유지되는 경우에만$$ X ${\displaystyle$ X$}$의 울트라필터입니다.^[2][3]

1. 모든 $S{\displaystyle }$ ${\displaystyle \subseteq }$ ${\displaystyle X}$, ${\displaystyle S\subseteq X,}$에 대해 S ${\displaystyle \in }$ ${\displaystyle U}$ ${\displaystyle$ S$\in U}$ 또는$$ $X{\displaystyle }$ ${\displaystyle \setminus }$ ${\displaystyle S.}$ { ${\displaystyle U.}$ ${\displaysty X\setminus$ S$\}$중 하나$.$
2. ${\displaystyle U}$ ${\displaystyle U}$은(는) $X{\displaystyle }$, ${\displaystyle$ X$}$에 있는 최대 필터 하위 베이스로, $F$ ${\displaystyle X}$에 있는 필터 $하위$ $베이스$라면 ${\displaystyle U}$⊆ F$${\$displaystyle$$U$\$subseteq F}$은 $U{\displaystyle }$= F$.}$을 $의미$한다.

$X{\displaystyle }$ ${\displaystyle X}$의 (property) 필터 $U{\displaystyle }$ ${\displaystyle U}$은(는) $X{\displaystyle }$ ${\displaystyle X}$의 울트라필터(으)로$$, 다음과 같은 동등한 조건 중 하나가 유지되는 경우에만$$ 해당된다.

1. ${\displaystyle U}$ ${\displaystyle U}$은(는) 매우 높음;
2. ${\displaystyle U}$ ${\displaystyle U}$은(는) 울트라 프리필터에 의해 생성되며$$,
3. ${\displaystyle \mathrm{모든}}$ 부분 집합 $S{\displaystyle }$ ${\displaystyle \subseteq }$ X${\displaystyle }$, ${\displaystyle$ S$\subseteq X,}$ $S$${\displaystyle }$ ${\displaystyle \in }$ ${\displaystyle U}$ ${\displaystyle$ S$\in U.}$ $또는$ X ${\displaystyle \setminus }$ ${\displaystyle S.}$S. { ${\displaystyle U.}$ ${\displaystyle X\setminus S\in.}$에 대하여
   • So an ultrafilter ${\displaystyle U}$ decides for every ${\displaystyle S\subseteq X}$ whether ${\displaystyle S}$ is "large" (i.e. ${\displaystyle S\in U}$) or "small" (i.e. ${\displaystyle X\setminus S\in U}$).^[7]
4. ${\displaystyle 각}$ 하위 집합 $A{\displaystyle }$ ${\displaystyle \subseteq }$ X${\displaystyle }$, ${\displaystyle A\subseteq X,}$에 대해 ${\displaystyle A}$이(가) $U{\displaystyle }$ ${\displaystyle U}$ 또는$$ ( ${\displaystyle X}$ ${\displaystyle \setminus }$ A ${\displaystyle$ X$\setminus$ A$\})$가^{[note 1]}().
5. ${\displaystyle U\cup (X\setminus$$}$ 이 조건은$$ 다음과 같이 재작성할 수 있다: $\wp {\displaystyle \mathrm{}}$( ${\displaystyle X}$) ${\displaystyle \wp (X)}$은(는) $U{\displaystyle }$ ${\displaystyle U}$과(와) 이중 $X{\displaystyle }$ ${\displaystyle \setminus }$ ${\displaystyle U.}$ ${\displaystyle X\setminus U.}$
   • $모든{\displaystyle }$ 프리필터 $P{\displaystyle }$ {\$displaystyle$ P}과$($와$$) X$$ P${\displaystyle }${\$displaystyle$ X$\setminus P}$ 세트는 X${\displaystyle }$. ${\displaystyle X.}$의 P ${\displaystystyle P}$에 대해 분리된다$$.
6. ${\$$S\in \wP(X):$$S\not \in U\$$right$$\}$은(는) $X{\displaystyle }$ . {\displaystyle $X.}$에 이상적이다.
7. For any finite family ${\displaystyle S_{1},\ldots ,S_{n}}$ of subsets of ${\displaystyle X}$ (where ${\displaystyle n\geq 1}$), if ${\displaystyle S_{1}\cup \cdots \cup S_{n}\in U}$ then ${\displaystyle S_{i}\in U}$ for some index ${\displays$$tyle i.}$
   • 즉, "큰" 집합은 크지 않은 집합의 유한 결합일 수 없다.^[8]
8. 모든 $R{\displaystyle }$, ${\displaystyle S}$ ${\displaystyle \subseteq }$ ${\displaystyle X}$, ${\displaystyle R,S\subseteq X}$의 $경우{\displaystyle }$, ${\displaystyle R}$ ${\displaystyle \cup }$ S${\displaystyle }$= ${\displaystyle X}$ ${\displaystyle R\cup S=X}$이면 $R{\displaystyle }$ ${\displaystyle \in }$ ${\displaystyle U}$ ${\displaystyle R\in U$${\displaystyle \}}$ 또는$$ $S{\displaystyle }$ ${\displaystyle \in }$ ${\displaystyle U}$. ${\displaystystyle S\in.}$
9. For any ${\displaystyle R,S\subseteq X,}$ if ${\displaystyle R\cup S\in U}$ then ${\displaystyle R\in U}$ or ${\displaystyle S\in U}$ (a filter with this property is called a prime filter).
10. For any ${\displaystyle R,S\subseteq X,}$ if ${\displaystyle R\cup S\in U}$ and ${\displaystyle R\cap S=\varnothing }$ then either ${\displaystyle R\in U}$ or ${\displaystyle S\in U.}$
11. ${\displaystyle U}$ is a maximal filter; that is, if ${\displaystyle F}$ is a filter on ${\displaystyle X}$ such that ${\displaystyle U\subseteq F}$ then ${\displaystyle U=F.}$ Equivalently, ${\displaystyle U}$ is a maximal filter if there is no filter ${\displaystyle F}$ on ${\displaystyle X}$$U{\displaystyle }$ ${\displaystyle U}$을(를) 적절한 하위 집합으로 포함하는$$$$ ${\displaystyle X}($즉, 필터는 $U{\displaystyle }$ ${\displaystyle U}$보다 엄격하게 미세하지 않음).$$^[4]

그릴 및 필터 그릴

$B{\displaystyle \mathcal{}}$ ${\displaystyle \mathrm{\subseteq (}X}$) ${\displaystyle {\mathcal$ {B$}\subseteq \$wp($X)}$인 경우 $X{\displaystyle }$ ${\displaystyle$ X$}$의 그릴이 패밀리임$$

$여기$서 $X{\displaystyle {\mathcal{}}^{}}$${\displaystyle$ ${\mathcal{B}^{\#}$는 문맥에서 명확하면$$ 쓸 수 있다$$.For example, ${\displaystyle \varnothing ^{\#}=\wp (X)}$ and if ${\displaystyle \varnothing \in {\mathcal {B}}}$ then ${\displaystyle {\mathcal {B}}^{\#}=\varnothing .}$ If ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {B}}}$ then ${\displaystyle {\mathcal{B}}^{\mathrm{\#}}\subseteq {\mathcal{A}}^{\mathrm{\#}}}${\displaystyle{{B\mathcal}}^{)#}\subseteq}{{A\mathcal}^{)#}}, 만약 B{\displaystyle{{B\mathcal}}}은 필터 subbase 그 다음에 B⊆ B#.{\displaystyle{{B\mathcal}}\subseteq}{{B\mathcal}^{)#}.}그 그릴 B#X{\displaystyle{{B\mathcal}}^{\)X}}상승 X에서{\d이 닫혀 있[9].$Isplaystyle X}$은${\displaystyle (}$는) assumed B${\displaystyle }$, ${\displaystyle \varnot \not \in {\mathcal {B},}$인 경우에만 해당되며, 이 경우에는$$ 다음이 가정된다.Moreover, ${\displaystyle {\mathcal {B}}^{\#\#}={\mathcal {B}}^{\uparrow X}}$ so that ${\displaystyle {\mathcal {B}}}$ is upward closed in ${\displaystyle X}$ if and only if ${\displaystyle {\mathcal {B}}^{\#\#}={\mathcal {B}}.$$}$

필터의 X{X\displaystyle}의 그릴 X.{X\displaystyle}[9]에 어떤 ∅ ≠ B⊆ ℘(X),{\varnothing \neq{\mathcal{B\displaystyle}}\subseteq\wp(X),}B{\displaystyle{{B\mathcal} 들어 filter-grill}}X{X\displaystyle}에 있는 filter-grill 만일(1)B{\disp라고 불린다.놓다$style {\mathcal {B}}}$ is upward closed in ${\displaystyle X}$ and (2) for all sets ${\displaystyle R}$ and ${\displaystyle S,}$ if ${\displaystyle R\cup S\in {\mathcal {B}}}$ then ${\displaystyle R\in {\mathcal {B}}}$ or ${\displaystyle S\in {\mathcal {B}}.}$ The grill 연산 $F{\displaystyle \mathcal{}}$ ${\displaystyle \mapsto }$ ${\displaystyle F^{}}$# X ${\$는 편차를 유도한다$$.

${\displaystyle {\bullet }^{\#X}~~:~\\operatorname {Filters}(X)\to \operatorname {FilterGrills}(X)}$

누구의 역 또한 F↦ F#X.{\displaystyle{{F\mathcal}}\mapsto{{F\mathcal}}^{\)X} 주어진다.X{X\displaystyle}에}[9]만약 F∈한 필터(X){\displaystyle{{F\mathcal}⁡}\in\operatorname{Filters}(X)} 다음 F{\displaystyle{{F\mathcal}}}은 filter-grill 만일 F)F.${\displaystyle {\mathcal {F}}={\mathcal {F}}^{\#X},}$^[9] or equivalently, if and only if ${\displaystyle {\mathcal {F}}}$ is an ultrafilter on ${\displaystyle X.}$^[9] That is, a filter on ${\displaystyle X}$ is a filter-grill if and only if it is ultra.For any non-empty ${\displaystyle {\mathcal {F}}\subseteq \wp (X),}$ ${\displaystyle {\mathcal {F}}}$ is both a filter on ${\displaystyle X}$ and a filter-grill on ${\displaystyle X}$ if and only if (1) ${\displaystyle \varnothing \not \in {\mathcal {F}}}$ and (2) for all ${\displaystyle R,}$${\displaystyle S}$ ${\displaystyle \subseteq }$ ${\displaystyle X}$, ${\displaystyle R,S\subseteq X}$ 등가물은 다음과 같다$$.

${\displaystyle R\cup S\in {\mathcal {F}}}$ if and only if ${\displaystyle R,S\in {\mathcal {F}}}$ if and only if ${\displaystyle R\cap S\in {\mathcal {F}}.}$^[9]

Free or principal

If ${\displaystyle P}$ is any non-empty family of sets then the Kernel of ${\displaystyle P}$ is the intersection of all sets in ${\displaystyle P:}$^[10]

A non-empty family of sets ${\displaystyle P}$ is called:

• free if ${\displaystyle \operatorname {ker} P=\varnothing }$ and fixed otherwise (that is, if ${\displaystyle \operatorname {ker} P\neq \varnothing }$).
• principal if ${\displaystyle \operatorname {ker} P\in P.}$
• principal at a point if ${\displaystyle \operatorname {ker} P\in P}$ and ${\displaystyle \operatorname {ker} P}$ is a singleton set; in this case, if ${\displaystyle \operatorname {ker} P=\{x\}}$ then ${\displaystyle P}$ is said to be principal at ${\displaystyle x.}$

If a family of sets ${\displaystyle P}$ is fixed then ${\displaystyle P}$ is ultra if and only if some element of ${\displaystyle P}$ is a singleton set, in which case ${\displaystyle P}$ will necessarily be a prefilter. Every principal prefilter is fixed, so a principal prefilter ${\displaystyle P}$ is ultra if and only if ${\displaystyle \operatorname {ker} P}$ is a singleton set. A singleton set is ultra if and only if its sole element is also a singleton set.

Every filter on ${\displaystyle X}$ that is principal at a single point is an ultrafilter, and if in addition ${\displaystyle X}$ is finite, then there are no ultrafilters on ${\displaystyle X}$ other than these.^[10] If there exists a free ultrafilter (or even filter subbase) on a set ${\displaystyle X}$ then ${\displaystyle X}$ must be infinite.

The next theorem shows that every ultrafilter falls into one of two categories: either it is free or else it is a principal filter generated by a single point.

Examples, properties, and sufficient conditions

If ${\displaystyle U}$ and ${\displaystyle S}$ are families of sets such that ${\displaystyle U}$ is ultra, ${\displaystyle \varnothing \not \in S,}$ and ${\displaystyle U\leq S,}$ then ${\displaystyle S}$ is necessarily ultra. A filter subbase ${\displaystyle U}$ that is not a prefilter cannot be ultra; but it is nevertheless still possible for the prefilter and filter generated by ${\displaystyle U}$ to be ultra.

Suppose ${\displaystyle U\subseteq \wp (X)}$ is ultra and ${\displaystyle Y}$ is a set. The trace ${\displaystyle U\cap Y:=\{B\cap Y:B\in U\}}$ is ultra if and only if it does not contain the empty set. Furthermore, at least one of the sets ${\displaystyle [U\cap Y]\setminus \{\varnothing \}}$ and ${\displaystyle [U\cap (X\setminus Y)]\setminus \{\varnothing \}}$ will be ultra (this result extends to any finite partition of ${\displaystyle X}$). If ${\displaystyle F_{1},\ldots ,F_{n}}$ are filters on ${\displaystyle X,}$ ${\displaystyle U}$ is an ultrafilter on ${\displaystyle X,}$ and ${\displaystyle F_{1}\cap \cdots \cap F_{n}\leq U,}$ then there is some ${\displaystyle F_{i}}$ that satisfies ${\displaystyle F_{i}\leq U.}$^[11] This result is not necessarily true for an infinite family of filters.^[11]

The image under a map ${\displaystyle f:X\to Y}$ of an ultra set ${\displaystyle U\subseteq \wp (X)}$ is again ultra and if ${\displaystyle U}$ is an ultra prefilter then so is ${\displaystyle f(U).}$ The property of being ultra is preserved under bijections. However, the preimage of an ultrafilter is not necessarily ultra, not even if the map is surjective. For example, if ${\displaystyle X}$ has more than one point and if the range of ${\displaystyle f:X\to Y}$ consists of a single point ${\displaystyle \{y\}}$ then ${\displaystyle \{y\}}$ is an ultra prefilter on ${\displaystyle Y}$ but its preimage is not ultra. Alternatively, if ${\displaystyle U}$ is a principal filter generated by a point in ${\displaystyle Y\setminus f(X)}$then the preimage of ${\displaystyle U}$ contains the empty set and so is not ultra.

The elementary filter induced by an infinite sequence, all of whose points are distinct, is not an ultrafilter.^[11] If ${\displaystyle n=2,}$ then ${\displaystyle U_{n}}$ denotes the set consisting all subsets of ${\displaystyle X}$ having cardinality ${\displaystyle n,}$ and if ${\displaystyle X}$contains at least ${\displaystyle 2n-1}$ (${\displaystyle =3}$) distinct points, then ${\displaystyle U_{n}}$ is ultra but it is not contained in any prefilter. This example generalizes to any integer ${\displaystyle n>1}$ and also to ${\displaystyle n=1}$ if ${\displaystyle X}$ contains more than one element. Ultra sets that are not also prefilters are rarely used.

For every ${\displaystyle S\subseteq X\times X}$ and every ${\displaystyle a\in X,}$ let ${\displaystyle S{\big \vert }_{\{a\}\times X}:=\left\{y\in X~:~(a,y)\in S\right\}.}$ If ${\displaystyle {\mathcal {U}}}$ is an ultrafilter on ${\displaystyle X}$ then the set of all ${\displaystyle S\subseteq X\times X}$ such that ${\displaystyle \left\{a\in X~:~S{\big \vert }_{\{a\}\times X}\in {\mathcal {U}}\right\}\in {\mathcal {U}}}$ is an ultrafilter on ${\displaystyle X\times X.}$^[12]

Monad structure

The functor associating to any set ${\displaystyle X}$ the set of ${\displaystyle U(X)}$ of all ultrafilters on ${\displaystyle X}$ forms a monad called the ultrafilter monad. The unit map

${\displaystyle X\to U(X)}$

sends any element ${\displaystyle x\in X}$ to the principal ultrafilter given by ${\displaystyle x.}$

This monad admits a conceptual explanation as the codensity monad of the inclusion of the category of finite sets into the category of all sets.^[13]

The ultrafilter lemma

The ultrafilter lemma was first proved by Alfred Tarski in 1930.^[12]

The ultrafilter lemma is equivalent to each of the following statements:

1. For every prefilter on a set ${\displaystyle X,}$ there exists a maximal prefilter on ${\displaystyle X}$ subordinate to it.^[2]
2. Every proper filter subbase on a set ${\displaystyle X}$ is contained in some ultrafilter on ${\displaystyle X.}$

A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.^{[14][note 2]}

The following results can be proven using the ultrafilter lemma. A free ultrafilter exists on a set ${\displaystyle X}$ if and only if ${\displaystyle X}$ is infinite. Every proper filter is equal to the intersection of all ultrafilters containing it.^[5] Since there are filters that are not ultra, this shows that the intersection of a family of ultrafilters need not be ultra. A family of sets ${\displaystyle \mathbb {F} \neq \varnothing }$ can be extended to a free ultrafilter if and only if the intersection of any finite family of elements of ${\displaystyle \mathbb {F} }$ is infinite.

Relationships to other statements under ZF

Throughout this section, ZF refers to Zermelo–Fraenkel set theory and ZFC refers to ZF with the Axiom of Choice (AC). The ultrafilter lemma is independent of ZF. That is, there exist models in which the axioms of ZF hold but the ultrafilter lemma does not. There also exist models of ZF in which every ultrafilter is necessarily principal.

Every filter that contains a singleton set is necessarily an ultrafilter and given ${\displaystyle x\in X,}$ the definition of the discrete ultrafilter ${\displaystyle \{S\subseteq X~:~x\in S\}}$ does not require more than ZF. If ${\displaystyle X}$ is finite then every ultrafilter is a discrete filter at a point; consequently, free ultrafilters can only exist on infinite sets. In particular, if ${\displaystyle X}$ is finite then the ultrafilter lemma can be proven from the axioms ZF. The existence of free ultrafilter on infinite sets can be proven if the axiom of choice is assumed. More generally, the ultrafilter lemma can be proven by using the axiom of choice, which in brief states that any Cartesian product of non-empty sets is non-empty. Under ZF, the axiom of choice is, in particular, equivalent to (a) Zorn's lemma, (b) Tychonoff's theorem, (c) the weak form of the vector basis theorem (which states that every vector space has a basis), (d) the strong form of the vector basis theorem, and other statements. However, the ultrafilter lemma is strictly weaker than the axiom of choice. While free ultrafilters can be proven to exist, it is not possible to construct an explicit example of a free ultrafilter; that is, free ultrafilters are intangible.^[15] Alfred Tarski proved that under ZFC, the cardinality of the set of all free ultrafilters on an infinite set ${\displaystyle X}$ is equal to the cardinality of ${\displaystyle \wp (\wp (X)),}$ where ${\displaystyle \wp (X)}$ denotes the power set of ${\displaystyle X.}$^[16] Other authors attribute this discovery to Bedřich Pospíšil (following a combinatorial argument from Fichtenholz, and Kantorovitch, improved by Hausdorff).^[17][18]

Under ZF, the axiom of choice can be used to prove both the ultrafilter lemma and the Krein–Milman theorem; conversely, under ZF, the ultrafilter lemma together with the Krein–Milman theorem can prove the axiom of choice.^[19]

Statements that cannot be deduced

The ultrafilter lemma is a relatively weak axiom. For example, each of the statements in the following list can not be deduced from ZF together with only the ultrafilter lemma:

1. A countable union of countable sets is a countable set.
2. The axiom of countable choice (ACC).
3. The axiom of dependent choice (ADC).

Equivalent statements

Under ZF, the ultrafilter lemma is equivalent to each of the following statements:^[20]

1. The Boolean prime ideal theorem (BPIT).
   • This equivalence is provable in ZF set theory without the Axiom of Choice (AC).
2. Stone's representation theorem for Boolean algebras.
3. Any product of Boolean spaces is a Boolean space.^[21]
4. Boolean Prime Ideal Existence Theorem: Every nondegenerate Boolean algebra has a prime ideal.^[22]
5. Tychonoff's theorem for Hausdorff spaces: Any product of compact Hausdorff spaces is compact.^[21]
6. If ${\displaystyle \{0,1\}}$ is endowed with the discrete topology then for any set ${\displaystyle I,}$ the product space ${\displaystyle \{0,1\}^{I}}$ is compact.^[21]
7. Each of the following versions of the Banach-Alaoglu theorem is equivalent to the ultrafilter lemma:
   1. Any equicontinuous set of scalar-valued maps on a topological vector space (TVS) is relatively compact in the weak-* topology (that is, it is contained in some weak-* compact set).^[23]
   2. The polar of any neighborhood of the origin in a TVS ${\displaystyle X}$ is a weak-* compact subset of its continuous dual space.^[23]
   3. The closed unit ball in the continuous dual space of any normed space is weak-* compact.^[23]
      • If the normed space is separable then the ultrafilter lemma is sufficient but not necessary to prove this statement.
8. A topological space ${\displaystyle X}$ is compact if every ultrafilter on ${\displaystyle X}$ converges to some limit.^[24]
9. A topological space ${\displaystyle X}$ is compact if and only if every ultrafilter on ${\displaystyle X}$ converges to some limit.^[24]
   • The addition of the words "and only if" is the only difference between this statement and the one immediately above it.
10. The Ultranet lemma: Every net has a universal subnet.^[25]
    • By definition, a net in ${\displaystyle X}$ is called an ultranet or an universal net if for every subset ${\displaystyle S\subseteq X,}$ the net is eventually in ${\displaystyle S}$ or in ${\displaystyle X\setminus S.}$
11. A topological space ${\displaystyle X}$ is compact if and only if every ultranet on ${\displaystyle X}$ converges to some limit.^[24]
    • If the words "and only if" are removed then the resulting statement remains equivalent to the ultrafilter lemma.^[24]
12. A convergence space ${\displaystyle X}$ is compact if every ultrafilter on ${\displaystyle X}$ converges.^[24]
13. A uniform space is compact if it is complete and totally bounded.^[24]
14. The Stone–Čech compactification Theorem.^[21]
15. Each of the following versions of the compactness theorem is equivalent to the ultrafilter lemma:
    1. If ${\displaystyle \Sigma }$ is a set of first-order sentences such that every finite subset of ${\displaystyle \Sigma }$ has a model, then ${\displaystyle \Sigma }$ has a model.^[26]
    2. If ${\displaystyle \Sigma }$ is a set of zero-order sentences such that every finite subset of ${\displaystyle \Sigma }$ has a model, then ${\displaystyle \Sigma }$ has a model.^[26]
16. The completeness theorem: If ${\displaystyle \Sigma }$ is a set of zero-order sentences that is syntactically consistent, then it has a model (that is, it is semantically consistent).

Weaker statements

Any statement that can be deduced from the ultrafilter lemma (together with ZF) is said to be weaker than the ultrafilter lemma. A weaker statement is said to be strictly weaker if under ZF, it is not equivalent to the ultrafilter lemma. Under ZF, the ultrafilter lemma implies each of the following statements:

1. The Axiom of Choice for Finite sets (ACF): Given ${\displaystyle I\neq \varnothing }$ and a family ${\displaystyle \left(X_{i}\right)_{i\in I}}$ of non-empty finite sets, their product ${\displaystyle \prod _{i\in I}X_{i}}$ is not empty.^[25]
2. A countable union of finite sets is a countable set.
   • However, ZF with the ultrafilter lemma is too weak to prove that a countable union of countable sets is a countable set.
3. The Hahn–Banach theorem.^[25]
   • In ZF, the Hahn–Banach theorem is strictly weaker than the ultrafilter lemma.
4. The Banach–Tarski paradox.
   • In fact, under ZF, the Banach–Tarski paradox can be deduced from the Hahn–Banach theorem,^[27][28] which is strictly weaker than the Ultrafilter Lemma.
5. Every set can be linearly ordered.
6. Every field has a unique algebraic closure.
7. The Alexander subbase theorem.^[25]
8. Non-trivial ultraproducts exist.
9. The weak ultrafilter theorem: A free ultrafilter exists on ${\displaystyle \mathbb {N} .}$
   • Under ZF, the weak ultrafilter theorem does not imply the ultrafilter lemma; that is, it is strictly weaker than the ultrafilter lemma.
10. There exists a free ultrafilter on every infinite set;
    • This statement is actually strictly weaker than the ultrafilter lemma.
    • ZF alone does not even imply that there exists a non-principal ultrafilter on some set.

Completeness

The completeness of an ultrafilter ${\displaystyle U}$ on a powerset is the smallest cardinal κ such that there are κ elements of ${\displaystyle U}$ whose intersection is not in ${\displaystyle U.}$ The definition of an ultrafilter implies that the completeness of any powerset ultrafilter is at least ${\displaystyle \aleph _{0}}$. An ultrafilter whose completeness is greater than ${\displaystyle \aleph _{0}}$—that is, the intersection of any countable collection of elements of ${\displaystyle U}$ is still in ${\displaystyle U}$—is called countably complete or σ-complete.

The completeness of a countably complete nonprincipal ultrafilter on a powerset is always a measurable cardinal.^{[citation needed]}

Ordering on ultrafilters

The Rudin–Keisler ordering (named after Mary Ellen Rudin and Howard Jerome Keisler) is a preorder on the class of powerset ultrafilters defined as follows: if ${\displaystyle U}$ is an ultrafilter on ${\displaystyle \wp (X),}$ and ${\displaystyle V}$ an ultrafilter on ${\displaystyle \wp (Y),}$ then ${\displaystyle V\leq {}_{RK}U}$ if there exists a function ${\displaystyle f:X\to Y}$ such that

${\displaystyle C\in V}$ if and only if ${\displaystyle f^{-1}[C]\in U}$

for every subset ${\displaystyle C\subseteq Y.}$

Ultrafilters ${\displaystyle U}$ and ${\displaystyle V}$ are called Rudin–Keisler equivalent, denoted U ≡_RK V, if there exist sets ${\displaystyle A\in U}$ and ${\displaystyle B\in V}$ and a bijection ${\displaystyle f:A\to B}$ that satisfies the condition above. (If ${\displaystyle X}$ and ${\displaystyle Y}$ have the same cardinality, the definition can be simplified by fixing ${\displaystyle A=X,}$ ${\displaystyle B=Y.}$)

It is known that ≡_RK is the kernel of ≤_RK, i.e., that U ≡_RK V if and only if ${\displaystyle U\leq {}_{RK}V}$ and ${\displaystyle V\leq {}_{RK}U.}$^[29]

Ultrafilters on ${\displaystyle \wp (\omega )}$

There are several special properties that an ultrafilter on ${\displaystyle \wp (\omega ),}$ where ${\displaystyle \omega }$ extends the natural numbers, may possess, which prove useful in various areas of set theory and topology.

• A non-principal ultrafilter ${\displaystyle U}$ is called a P-point (or weakly selective) if for every partition ${\displaystyle \left\{C_{n}:n<\omega \right\}}$ of ${\displaystyle \omega }$ such that for all ${\displaystyle n<\omega ,}$ ${\displaystyle C_{n}\not \in U,}$ there exists some ${\displaystyle A\in U}$ such that ${\displaystyle A\cap C_{n}}$ is a finite set for each ${\displaystyle n.}$
• A non-principal ultrafilter ${\displaystyle U}$ is called Ramsey (or selective) if for every partition ${\displaystyle \left\{C_{n}:n<\omega \right\}}$ of ${\displaystyle \omega }$ such that for all ${\displaystyle n<\omega ,}$ ${\displaystyle C_{n}\not \in U,}$ there exists some ${\displaystyle A\in U}$ such that ${\displaystyle A\cap C_{n}}$ is a singleton set for each ${\displaystyle n.}$

It is a trivial observation that all Ramsey ultrafilters are P-points. Walter Rudin proved that the continuum hypothesis implies the existence of Ramsey ultrafilters.^[30] In fact, many hypotheses imply the existence of Ramsey ultrafilters, including Martin's axiom. Saharon Shelah later showed that it is consistent that there are no P-point ultrafilters.^[31] Therefore, the existence of these types of ultrafilters is independent of ZFC.

P-points are called as such because they are topological P-points in the usual topology of the space βω \ ω of non-principal ultrafilters. The name Ramsey comes from Ramsey's theorem. To see why, one can prove that an ultrafilter is Ramsey if and only if for every 2-coloring of ${\displaystyle [\omega ]^{2}}$ there exists an element of the ultrafilter that has a homogeneous color.

An ultrafilter on ${\displaystyle \wp (\omega )}$ is Ramsey if and only if it is minimal in the Rudin–Keisler ordering of non-principal powerset ultrafilters.^[32]

See also

• Filter (mathematics) – In mathematics, a special subset of a partially ordered set
• Filters in topology – Use of filters to describe and characterize all basic topological notions and results.
• Universal net

Notes

Proofs

References

Bibliography

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Further reading

• Comfort, W. W. (1977). "Ultrafilters: some old and some new results". Bulletin of the American Mathematical Society. 83 (4): 417–455. doi:10.1090/S0002-9904-1977-14316-4. ISSN 0002-9904. MR 0454893.
• Comfort, W. W.; Negrepontis, S. (1974), The theory of ultrafilters, Berlin, New York: Springer-Verlag, MR 0396267
• Ultrafilter in nLab