3
$\begingroup$

My confusion concerns ultrafilters on sets that are themselves power sets.

If $X=\{\emptyset,\ \{1\},\ \{2\},\ \{3\},\ \{4\},\ \{1,2\},\ \{1,3\},\ \{1,4\},\ \{2,3\},\ \{2,4\},\ \{3,4\},\ \{1,2,3\},\ \{1,2,4\},\ \{1,3,4\},\ \{2,3,4\},\ \{1,2,3,4\}\ \}$ and the upset $\{1\}=U=\{\ \{1\},\ \{1,2\},\ \{1,3\},\ \{1,4\},\ \{1,2,3\},\ \{1,2,4\},\ \{1,3,4\},\ \{1,2,3,4\}\ \}$ is supposedly a principal ultrafilter (for visual delineation see https://en.wikipedia.org/wiki/Filter_%28mathematics%29), then how do I satisfy the criteria that "If $A$ is a subset of $X$, then either $A$ or $X\setminus A$ is an element of $U$"?

For example, could I let $A=\{\emptyset,\{2\}\}$ such that $A\notin U$, but $X\setminus A\notin U$ because $\{2,3,4\}\in X\setminus A,\{2,3,4\}\notin U$?

My understanding of the distinction between an element and a subset is unrefined, particularly with regard to power sets.

$\endgroup$
2
  • 2
    $\begingroup$ $U$ is a principal ultrafilter on $M = \{1,2,3,4\}$, not on $\mathfrak{P}(M)$. $\endgroup$ Commented Jul 2, 2013 at 21:52
  • $\begingroup$ That hit the nail on the head, thank you $\endgroup$
    – user84815
    Commented Jul 2, 2013 at 22:02

2 Answers 2

5
$\begingroup$

You’ve completely confused elements and subsets, I’m afraid. $X$ here is the set $\{1,2,3,4\}$, and what you called $X$ is actually $\wp(X)$, the power set of $X$, i.e., the set of all subsets of $X$. Your $A$ is not a subset of $X$; it’s a subset of $\wp(X)$. A couple of subsets of $X$ are $A_1=\{1,3\}$ and $A_2\{2,4\}$. As you can see, $A_1\in U$; $A_2\notin U$, but $X\setminus A_2=A_1$ is an element of $U$.

$\endgroup$
2
  • $\begingroup$ That makes perfect sense. I think I led myself astray by assuming that the author was claiming that upset in question was a principal ultrafilter of ℘(X) and not X. Thank you. $\endgroup$
    – user84815
    Commented Jul 2, 2013 at 22:06
  • $\begingroup$ @user84815: You’re welcome. $\endgroup$ Commented Jul 2, 2013 at 22:11
2
$\begingroup$

It is fine to have filters on a power set of another set. But then the filter is a subset of the power set of the power set of that another set; rather than our original set which is the power set of another set.

But in the example that you gave, note that $W\in U$ if and only if $\{1\}\in W$. Indeed $\{1\}\notin A$, but therefore it is in its complement and so $X\setminus A\in U$.

If things like that confuse you, just replace the sets with other elements, $1,\ldots,16$ for example, and consider the ultrafilter on that set. Then simply translate it back to your original $X$.

$\endgroup$
2
  • $\begingroup$ Thank you for the response. I will employ your method for avoiding confusion. $\endgroup$
    – user84815
    Commented Jul 2, 2013 at 22:00
  • 2
    $\begingroup$ @user84815: I suggest this method whenever there is a construction invariant of extra structure on the set (e.g. filters, power sets, etc.) and the nature of the elements is confusing. For example in the question whether or not $0\in\{\{0\}\}$, we can replace $\{0\}$ by $x$ then we ask $0\in\{x\}$ and the answer is true if and only if $0=x$, but clearly $0\neq\{0\}$, so $0\notin\{x\}=\{\{0\}\}$. $\endgroup$
    – Asaf Karagila
    Commented Jul 2, 2013 at 22:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .