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A filter $\mathcal F$ is called free if $\bigcap \mathcal F=\emptyset$. Filter, which is not free is called principal. any principal ultrafilter has the form $$\mathcal F_a=\{A\subseteq X; a\in A\}$$ for some $a\in X$. Fréchet filter (or. cofinite filter) on a set $X$ is the filter consisting of all cofinite set, i.e., it is equal to $$\mathcal F_{F}=\{A\subseteq X; X\setminus A\text{ is finite}\}.$$

Is an ultrafilter free if and only if it contains the cofinite filter ? Why?

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Yes. Let $\mathscr{U}$ be an ultrafilter on $X$, and let $a\in X$ be arbitrary. Since $\mathscr{U}$ is an ultrafilter, either $\{a\}\in\mathscr{U}$, or $X\setminus\{a\}\in\mathscr{U}$. If $\{a\}\in\mathscr{U}$, then $\mathscr{F}_a\subseteq\mathscr{U}$, and therefore $\mathscr{U}=\mathscr{F}_a$. If $X\setminus\{a\}\in\mathscr{U}$ for each $a\in X$, then $X\setminus F\in\mathscr{U}$ for each finite $F\subseteq X$; this is easily proved by induction on $|F|$.

Alternatively, you can argue as follows. Let $\mathscr{F}$ be the cofinite filter on $X$. If $\mathscr{U}\supseteq\mathscr{F}$, then $$\bigcap\mathscr{U}\subseteq\bigcap\mathscr{F}=\varnothing\;,$$ so $\mathscr{U}$ cannot be a principal filter: $\bigcap\mathscr{F}_a=\{a\}$ for each $a\in X$. If, on the other hand, $\mathscr{U}\nsupseteq\mathscr{F}$, then there is a finite $F\subseteq X$ such that $X\setminus F\notin\mathscr{U}$. Let $F=\{x_1,\ldots,x_n\}$; clearly $$F=\{x_1\}\cup\{x_2\}\cup\ldots\cup\{x_n\}\;,$$ a finite union, so there is exactly one $k\in\{1,\dots,n\}$ such that $\{x_k\}\in\mathscr{U}$, and it follows immediately that $\mathscr{U}=\mathscr{F}_{x_k}$, the principal ultrafilter over $x_k$.

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  • $\begingroup$ can we say every ultrafilter on infinite set is free?(2)what is difference between the meaning a principal filter and ultrafilter? $\endgroup$ – fatemeh Sep 13 '13 at 20:41
  • $\begingroup$ @fatemeh: (1) No, of course not: if $X$ is an infinite set, and $x\in X$, then $\{A\subseteq X:x\in A\}$ is a principal ultrafilter on $X$, not a free ultrafilter. (2) If $S$ is a non-empty subset of $X$, then $\{A\subseteq X:S\subseteq A\}$ is a principal filter on $X$; specifically, it’s the principal filter over the set $S$. If $S=\{x\}$ for some $x\in X$, then one can prove that this principal filter is actually an ultrafilter — specifically, a principal ultrafilter. The principal ultrafilters on $X$ are precisely the principal filters that are also ultrafilters. The filters and ... $\endgroup$ – Brian M. Scott Sep 13 '13 at 20:57
  • $\begingroup$ ... ultrafilters on $X$ that are not principal are the free filters and ultrafilters; they all contain the cofinite filter. $\endgroup$ – Brian M. Scott Sep 13 '13 at 20:57
  • $\begingroup$ Is ihe same difinition of limit in filter and ultrafilter? how does limit of ultrafilter define a topology on X? $\endgroup$ – fatemeh Sep 13 '13 at 21:06
  • $\begingroup$ @fatemeh: If $\langle X,\tau\rangle$ is a topological space, $\mathscr{F}$ is a filter on $X$, and $x\in X$, then $\mathscr{F}\to x$ iff every nbhd of $x$ belongs to $\mathscr{F}$. This definition of filter convergence applies to all filter, including ultrafilters. I’m not sure what your second question means. One normally talks about convergence of filters when one already has some topology on the underlying set. $\endgroup$ – Brian M. Scott Sep 13 '13 at 21:12
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Remember that a filter $\mathcal{F}$ is an ultrafilter if and only if it has the property

$$\bigl(\forall A \subset X\bigr)\bigl( A \in \mathcal{F} \lor (X\setminus A)\in\mathcal{F}\bigr).$$

An ultrafilter is principal if and only if it contains a finite set. So by the above characterisation, an ultrafilter is not principal if and only if it contains the complements of all finite sets, i.e. if and only if it contains the Fréchet filter.

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  • $\begingroup$ Is definition of uniform ultrafilter, cardinality of every element is infinite? or not $\endgroup$ – fatemeh Sep 13 '13 at 21:03
  • $\begingroup$ No it's not part of any definition. But a filter that contains a finite set must be fixed, $\bigcap \mathcal{F} \neq \varnothing$. For if $A \in \mathcal {F}$ is finite, and for each $a\in A$ there is an $F_a \in \mathcal{F}$ with $a \notin F_a$, then $A \cap \bigcap\limits_{a\in A} F_a = \varnothing$ is a finite intersection of elements of $\mathcal{F}$. So $\mathcal{F}$ is not a filter. $\endgroup$ – Daniel Fischer Sep 13 '13 at 21:09
  • $\begingroup$ @fatemeh: If $\mathscr{U}$ is a uniform ultrafilter on an infinite set $X$, then it’s true that every element of $\mathscr{U}$ is infinite, but this is not part of the definition: it’s an immediate consequence of the definition, which is that $|U|=|X|$ for all $U\in\mathscr{U}$. It’s also true that an ultrafilter $\mathscr{U}$ on an infinite set $X$ is free iff every $U\in\mathscr{U}$ is infinite, but if $X$ is uncountable, $\mathscr{U}$ can be free but not uniform. $\endgroup$ – Brian M. Scott Sep 13 '13 at 21:15
  • $\begingroup$ could you give me exact definition of uniform ultrafilter ?Thanks. $\endgroup$ – fatemeh Sep 13 '13 at 21:36
  • $\begingroup$ @fatemeh: An ultrafilter $\mathscr{U}$ on a set $X$ is uniform if $|U|=|X|$ for every $U\in\mathscr{U}$. $\endgroup$ – Brian M. Scott Sep 16 '13 at 0:46

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