# Tag Info

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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 ...
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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 ...
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### 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 ...
• 394k
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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. ...
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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 ...
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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 ...

### 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 ...
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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 ...
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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 $Y=X\sqcup\{a,b\}$ ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 $X\notin F$ ...
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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 ...
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### 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 ...
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### 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-...
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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 $V_\kappa$. ...
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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 ...