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I just started trying to read through Lectures on the Hyperreals: An Introduction to Nonstandard Analysis, by Robert Goldblatt. I'm near the beginning in the section on filters. He's defined an ultrafilter as follows: Consider a non-empty set $I$ and it's Power Set $\mathcal P(I)$. An ultrafilter $\mathcal F$ on $I$. contains all supersets of its elements as well as all finite intersections. Also $A\in$ $\mathcal F$ iff $A^c\notin\mathcal F$. Where $A^c = I-A$ and $A\in \mathcal P(I)$.

He then goes on to give some examples of filters, including 2 that are examples of "principal ultrafilters", however he doesn't actually define principal ultrafilters.

The first example would be $\mathcal F^i=\{A\subseteq I:i \in A \}$.

The second example is similar, but for a set, rather than a single element. $\mathcal F^\mathcal H=\{A\subseteq I:A \supseteq B \}$ and $B\ \in \mathcal P(I)$.

I'm unsure of the actual definition of a principal ultrafilter. It appears that it's a filter generated by an object (via supersets), where that object is either an element of the universe or an element of the power set of the universe. Along with satisfying the requirements of being an ultrafilter. Is that correct?

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    $\begingroup$ Are you sure that Goldblatt doesn't define principal ultrafilters as the filters $\mathcal{F}^i$? $\endgroup$ Jun 7, 2014 at 14:57
  • $\begingroup$ For fixed $a$ in $I$, it is the collection of subsets of $I$ that contain $a$. I think of an ultrafilter as a (finitely additive) "measure" on the power set, with values $0$ or $1$, that gives mass $1$ to $I$. A principal ultrafilter assigns mass $1$ to a certain point $a$. Non-principal ultrafilters assign mass $0$ to all points. $\endgroup$ Jun 7, 2014 at 15:06
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    $\begingroup$ @MichaelAlbanese - he does say that it's called the principal ultrafilter generated by i. But then he goes on to do the same thing with a element of the power set B. Calling this a special case of the situation when B={i}. $\endgroup$ Jun 7, 2014 at 15:29
  • $\begingroup$ But does he call such things principal ultrafilters? $\endgroup$ Jun 7, 2014 at 15:39
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    $\begingroup$ @MichaelAlbanese - yikes - you're right, he did not. He called them Principal Filters. Maybe I need to enroll in a reading class! Thank you so much! $\endgroup$ Jun 7, 2014 at 15:57

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Principal filters are filters which has a $\subseteq$ minimum, or in other words $\cal F$ is a principal filter if $\bigcap\cal F\in\cal F$. So if $\cal F$ is a principal filter and $A=\bigcap\cal F$ then $\mathcal F=\{B\subseteq X\mid A\subseteq B\}$.

If $\cal F$ is a principal ultrafilter then we can prove that there is a singleton in $\cal F$, which has to be unique. If this singleton is $\{i\}$ then $\mathcal F=\{A\subseteq X\mid\{i\}\subseteq A\}=\{A\subseteq X\mid i\in A\}$.

The proof is not difficult, if $A=\bigcap\cal F$ and $A$ is not a singleton, then $A$ is the disjoint union of $B\cup C$ where neither is empty. Since $A$ is the minimum of $\cal F$ neither $B$ nor $C$ are in $\cal F$; and since it's an ultrafilter $X\setminus B\in\cal F$, and therefore $A\cap(X\setminus B)=C\in\cal F$ which is a contradiction.

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    $\begingroup$ Just a note on something that confused me at first because I thought, "filters are only closed under FINITE intersections, not arbitrary ones". A filter $\mathcal{F}$ is of the form $\{ B \subseteq X \mid A \subseteq B \}$ for some $A \subseteq X$ iff $\bigcap \mathcal{F} \in \mathcal{F}$. Left to right: $\bigcap \mathcal{F} = A$. Right to left: $\bigcap \mathcal{F} \subseteq B$ for all $B \in \mathcal{F}$. $\endgroup$ Jul 16, 2021 at 16:40
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    $\begingroup$ @Daniel: $X$ is just the set on which the filter is defined. Your comments are a bit hard to understand since you're putting unnecessary quantifiers in some places. But if I understand correctly, then yes, $D$ and $F$ are exactly the same filter, with just different letters. This is like saying that $|x|<1\to x^2<1$ is hard to understand, because the book you're looking it up in says $0<r<1\to r\cdot r<1$. $\endgroup$
    – Asaf Karagila
    Aug 8, 2021 at 11:18
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    $\begingroup$ @DanielMak: Prove that if $A\in F$, then $\bigcap F\subseteq A$. Which is fairly straightforward. $\endgroup$
    – Asaf Karagila
    Aug 10, 2021 at 15:24
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    $\begingroup$ @Daniel: No. You assume that $\bigcap F\in F$, and you want to show that there is some set in $F$ which is a subset of every other element of $F$, in other words, that $F$ has a minimum with respect to inclusion. You're making too quick jumps, and you fall. I've been there, and I know it's hard, but work slowly with the definitions. Read this: karagila.org/2015/how-to-solve-your-problems $\endgroup$
    – Asaf Karagila
    Aug 11, 2021 at 6:40
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    $\begingroup$ @Daniel: Yes. This can be written better and more clearly (the use of letters is a bit confusing), but the argument is fine. $\endgroup$
    – Asaf Karagila
    Aug 12, 2021 at 11:54
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Goldblatt writes on page 18:

$\mathcal F^i=\{A\subseteq I; i\in A\}$ is an ultrafilter, called the principal ultrafilter generated by $i$. If $I$ is finite, then every ultrafilter on $I$ is of the form $\mathcal F^i$.

I think this passage can be considered the definition of principal ultrafilters.

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Let $X$ be any set and $x \in X$ then a principal ultrafilter is of the form

$$\mathcal{U}=\{ A \subseteq X | x\in A\}.$$

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    $\begingroup$ why the downvote ? $\endgroup$ Jun 7, 2014 at 16:43

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