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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 $\mathcal{F}^i$? $\endgroup$ – Michael Albanese Jun 7 '14 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. $\endgroup$ – André Nicolas Jun 7 '14 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}. $\endgroup$ – Mitchell Kaplan Jun 7 '14 at 15:29
  • $\begingroup$ But does he call such things principal ultrafilters? $\endgroup$ – Michael Albanese Jun 7 '14 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! $\endgroup$ – Mitchell Kaplan Jun 7 '14 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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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 ? $\endgroup$ – Rene Schipperus Jun 7 '14 at 16:43

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