17 votes
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Intuition for the Stone-Čech compactification via ultrafilters

Since you already know about the Alexandroff one-point compactification, let me begin by saying that the Stone-Cech compactification is at the other extreme, adding as many points at infinity as ...
Andreas Blass's user avatar
16 votes
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Example of a free ultrafilter on natural numbers

I'm assuming you meant "free ultrafilter." This is a very reasonable question! Unfortunately, in a very real sense we can't exhibit a concrete example of a free ultrafilter on $\mathbb{N}$ - it is ...
Noah Schweber's user avatar
12 votes

Can you get a non-principal ultrafilter on N using Choice but 'avoiding' Zorn's Lemma?

Of course. Fix a choice function on $\mathcal{P(P(\Bbb N))}$. Start with a non-principal filter, say the co-finite filter. By transfinite induction, each time choose a set not in the filter ...
Asaf Karagila's user avatar
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12 votes
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Meaning behind Filter in Set Theory

When you put a filter in your sink, the idea is that you filter out the big chunks of food, and let the water and the smaller chunks (which can—in principle—be washed through the pipes) go through. ...
Asaf Karagila's user avatar
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12 votes
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What happens to the Stone-Cech compactification if you change "compact Hausdorff" to "compact"?

No, if you drop the Hausdorff condition when talking about the Stone-Cech compactification, then it never exists for any non-compact space. Indeed, suppose $X$ is not compact and suppose there ...
Eric Wofsey's user avatar
11 votes
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Does the term "free" in "free ultrafilter" have a meaning related to category theory?

That's unlikely since free ultrafilters are not unique, but universal properties always imply a kind of uniqueness. MO/410462 has more information how different free ultrafilters can be (check also ...
Martin Brandenburg's user avatar
10 votes

In a finite lattice, every filter is principal.

This is almost correct, but you haven't explained why $\bigwedge F\in F$. Just because $\bigwedge F$ exists in $L$ does not automatically mean it must be an element of $F$. You need to again use the ...
Eric Wofsey's user avatar
10 votes
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In the ultrafilter lemma, is the ultrafilter unique?

No, it isn't even vaguely unique. For any set $X$, the set $\{X\}$ is a filter; this filter is not just contained in several ultrafilters, it's contained in all of them. So, for example, let $X$ be $\...
Reese Johnston's user avatar
10 votes
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Can unlimited hypernaturals be represented by increasing sequences?

You certainly can't make $(a_n)$ be strictly monotonic, since that would mean that $a_n\geq a_0+n$ for all $n$. So, for instance, if $k_n=\lfloor\sqrt{n}\rfloor$ we cannot choose any such $a_n$, ...
Eric Wofsey's user avatar
9 votes
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free ultrafilters on $\mathbb{N}$

We can't have disjoint ultrafilters, since every ultrafilter contains $\mathbb{N}$. Counting the nonprincipal ultrafilters takes a bit more work. It's easy to show that there are at most $2^{2^{\...
Noah Schweber's user avatar
9 votes
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Nonisomorphic free ultrafilters on $\omega$

The simplest property that I can think of (right now) that provably (in ZFC) distinguishes some non-principal ultrafilters on $\mathbb N$ from others is "weak P-point", which means "not ...
Andreas Blass's user avatar
9 votes
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Is the characterization of Hausdorff spaces in terms of ultrafilter convergence equivalent to the ultrafilter lemma?

Yes, this is equivalent to the ultrafilter lemma. Let $F$ be a proper filter on a set $X$, which we may assume to not be contained in any principal ultrafilter. Consider the space $Y=X\sqcup\{a,b\}$ ...
Eric Wofsey's user avatar
8 votes
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Another Topology on the Prime Spectrum of a Ring

This is the constructible topology, the topology generated by both the Zariski open sets and the basic Zariski closed sets (i.e., the closed sets defined by principal ideals). Another way to say it ...
Eric Wofsey's user avatar
7 votes
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Intuition about filters

Filters are ubiquitious in logic, where they are used to construct a single model from a bunch of models (a form of product) and in topology, where they can be used to speak of convergence in all ...
Ittay Weiss's user avatar
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7 votes
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"All ultrafilters are principal" consistent with ZF?

It is easy to construct filters which have no principal ultrafilters which extend them. For example, any filter extending the cofinite filter. But in order to prove that there is any filter ...
Asaf Karagila's user avatar
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7 votes
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What is the cardinality of the set of all ultrafilters containing a Fréchet filter?

Yes. If an ultrafilter $U$ does not contain the Fréchet filter, then $U$ must be principal. A principal ultrafilter is determined by its generating element, so there are $|X|$-many principal ...
Alex Kruckman's user avatar
7 votes
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Two topologies are equal if they have the same filter convergence

This is true. To see why just recall that $x\in \overline{A} \iff$ there is an ultrafilter $\mathcal{F}\rightarrow x$ with $A\in \mathcal{F}$. And show the closures of any subset must be identical ...
fosho's user avatar
  • 6,344
7 votes
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Filter that's neither Principal nor Ultrafilter

The Fréchet filter $F$ isn't the smallest non-principal ultrafilter. It's the filter consisting of all cofinite sets. And it's not an ultrafilter, since any infinite/co-infinite $X$ has $X\notin F$ ...
spaceisdarkgreen's user avatar
7 votes
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Is $\mathbb{Q}\;\cong\; (\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$?

This is a simple cardinality argument. Ultraproducts of finite sets are either finite or uncountable. Since the ultrafilter is free, and the sets are all increasing in size, it is not finite. To see ...
Asaf Karagila's user avatar
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7 votes

Is $\mathbb{Q}\;\cong\; (\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$?

Let's call your ultraproduct $K$. What's clear is that $K$ is a field of characteristic $0$ (by Łoś's theorem) of cardinality $2^{\aleph_0}$ (by the argument in Asaf's answer). What's less clear is ...
Alex Kruckman's user avatar
7 votes
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Lindelöf in terms of filters

A space $X$ is Lindelöf iff every filter on $X$ with the countable intersection property has a cluster point. (A filter $\mathscr{F}$ has the countable intersection property if $\bigcap\mathscr{C}\ne\...
Brian M. Scott's user avatar
7 votes
Accepted

Translation invariant ultrafilters?

Let $E$ and $O$ be the set of even and odd naturals respectively, and note that $E-1=O$ and $O-1=E$. If $\mathcal{U}$ is an ultrafilter, either $E\in\mathcal{U}$ and $O\not\in\mathcal{U}$ or $E\not\in\...
Noah Schweber's user avatar
7 votes

Localization and Field

Denote $R=k^I$ (I used the letter $A$ a lot in my answer for subsets of $I$, so should not use it to denote the ring). Consider the elements $e_A\in k^I$ for $A\subseteq I$, which have $i$'th co-...
tkf's user avatar
  • 11.5k
7 votes
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Is every "filter" of rings principal?

To avoid technicalities about what exactly you mean by "class", let me assume you are working in a Grothendieck universe $V_\kappa$ and "class" means any subset of $V_\kappa$. ...
Eric Wofsey's user avatar
6 votes

Condition for an Ultrafilter to be Ramsey.

I told Asaf that I'd write out the proof, so here goes. I'll divide it into individual steps that use different parts of the hypothesis. Assume that $\mathcal D$ is an ultrafilter on $\omega$ such ...
Andreas Blass's user avatar
6 votes
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Every net has an ultranet as subnet: direct proof

Let $f: I \to X$ be net in $X$, where $(I, \le)$ is a directed set. To build a universal subnet: let $\mathcal{F}$ be the filter of tails on $X$, so the filter generated by all $T(i) = \{f(j): j \ge i\...
Henno Brandsma's user avatar
6 votes
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Why do ultraproduct structures use a quotient as their universe?

For the most part, model theorists are concerned with complete theories. This means, for a fixed language $\mathcal{L}$, an $\mathcal{L}-$theory $T$ is complete if for every $\varphi$ in this language,...
Kyle Gannon's user avatar
  • 6,363
6 votes
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Non-principal ultrafilters on a set

There is no explicit example of a non-principal ultrafilter over the natural numbers without appealing to choice. We know this because there are models of $\sf ZF$ where every ultrafilter over the ...
Asaf Karagila's user avatar
  • 394k

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