39 votes
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Are all free ultrafilters 'the same' in some sense?

Certain important properties are shared by all free ultrafilters. In many applications of ultrafilters, especially more elementary applications, only these properties are used. In such a situation, it ...
Will Brian's user avatar
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23 votes
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Continuous images of $\beta \mathbb{N} \setminus\mathbb{N}$

This is a great question. There has been quite a bit of work done to figure out what the continuous images of $\beta \mathbb N \setminus \mathbb N$ are. I'll do my best to summarize some of that work ...
Will Brian's user avatar
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17 votes

Is $\beta \mathbb{N}$ homeomorphic to its own square?

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic. To derive a contradiction, assume that $\beta\mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are ...
Taras Banakh's user avatar
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16 votes

Near permutation $n\mapsto n+1$ not conjugate to its inverse on the Stone-Čech remainder?

Update: The answer is yes -- if $\mathsf{CH}$ is true then $\phi$ and $\phi^{-1}$ are conjugate in the group of self-homeomorphisms of $\omega^*$. I've written this up in a new paper, which you can ...
Will Brian's user avatar
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16 votes
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Is this space the Stone–Čech compactification?

No, the closure of the image of $f$ in $Y$ is never the Stone-Čech compactification of $X$ unless $X$ is empty. In particular, consider the element $a\in Y$ which is $1$ on every coordinate. Note ...
Eric Wofsey's user avatar
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13 votes

Is $\beta \mathbb{N}$ homeomorphic to its own square?

(Just noticed - already done by Todd Trimble in a comment:) A proof by Stone duality: the dual question is whether the Boolean algebras $\mathscr P\mathbb N$ and $\mathscr P\mathbb N\otimes\mathscr P\...
მამუკა ჯიბლაძე's user avatar
10 votes

Is $\beta \mathbb{N}$ homeomorphic to its own square?

The negative answer is equivalent to showing that there are two disjoint subsets $A,B$ of $\mathbf{N}^2$ with non-disjoint closures in $(\beta\mathbf{N})^2$. This be made explicit: take $A=\{(n,m):n=m\...
YCor's user avatar
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10 votes

The need for nets in topology

There seems to be a gap in your argument where, under the assumption that the system of neighborhoods could be totally ordered, you seem to use that this total order must be countable. To fill the ...
Andreas Blass's user avatar
10 votes
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Characterization of pretty compact spaces

A partial answer: other examples of pretty compact spaces are uncountable powers of $\{0,1\}$ and $[0,1]$, and in general products of uncountably many non-trivial compact Hausdorff spaces. See Problem ...
KP Hart's user avatar
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10 votes

Are all free ultrafilters 'the same' in some sense?

"Fast" ultrafilters and "slow" ultrafilters are used in this paper and have different properties. The definition of "slow" ultrafilter: let $\mathcal P$ be the set of ...
markvs's user avatar
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9 votes
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Addition and Rudin-Keisler ordering in $\beta \omega$

For Question 1: First, there are idempotent ultrafilters, so some ultrafilters satisfy 1 in a very strong form. But 1 does not hold in general. The reason is that the semigroup $\beta\omega-\omega$ ...
Andreas Blass's user avatar
9 votes
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Stone–Čech compactification as a semigroup

Corollary 4.33 of Hindman and Strauss's book on Algebra in the Stone Cech Compactification says that if $S$ is an infinite cancellative (discrete) semigroup, then the nonprincipal ultrafilters in $\...
Benjamin Steinberg's user avatar
8 votes
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Sigma algebras on the Stone–Čech compactification of a countable discrete group

No. First of all, take note that for compact zero-dimensional spaces $X$, the $\sigma$-algebra generated by all clopen sets is precisely the $\sigma$-algebra of Baire sets (recall that in a ...
Joseph Van Name's user avatar
8 votes

Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology

Topologically speaking $\mathbb{Z}$ with the topology mentioned above is just (homeomorphic to) the space of rational numbers. The space $\beta\mathbb{Q}$ has been studied a lot (not as much as $\beta\...
KP Hart's user avatar
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8 votes

Are all free ultrafilters 'the same' in some sense?

It is consistent that all ultrafilters are the same in the following sense: for any free ultrafilters $\mathcal U,\mathcal V$ on $\omega$ there exists a finite-to-one map $f:\omega\to\omega$ such that ...
Taras Banakh's user avatar
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8 votes

Is this space the Stone–Čech compactification?

Correction (2022-01-01): In the first version of this answer, I had defined $E$-compact spaces as those $X$ such that the natural evaluation map $X \to E^{C(X,E)}$ is a homeomorphism on a closed ...
Gro-Tsen's user avatar
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7 votes
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Continuous binary operations on $\beta\mathbb{N}$

In this paper, Dimension phenomena associated with $\beta\mathbb{N}$-spaces, Ilijas Farah proved that continuous maps from $\beta\mathbb{N}^2$ (and other powers) to $\beta\mathbb{N}$ are quite simple: ...
KP Hart's user avatar
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7 votes

The Stone-Čech compactification of a inverse system

Let $X_n$ be $\{k\in\mathbb{N}:k\ge n\}$ and let $f_n:X_{n+1}\to X_n$ be the inclusion map. The inverse limit of the system $\{X_n,f_n,\mathbb{N}\}$ is empty; the limit of the system $\{\beta X_n,\...
KP Hart's user avatar
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7 votes
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When are the zero sets of two continuous functions in the Stone-Čech compactification included in one another?

Lemma: Suppose that $X$ is a compact Hausdorff space. Let $f,g:X\rightarrow[0,\infty)$ be continuous functions. Then the following are equivalent: $Z(f)\subseteq Z(g)$. For all $\epsilon>0$, ...
Joseph Van Name's user avatar
7 votes
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Are $\beta \mathbb{Q}$ and $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ homeomorphic?

Let me summarize the discussion in the comments as an answer. Let $\chi(x, Y)$ be the character of $x$ in $Y$ i.e. the least cardinality of a local basis of the point $x$ in space $Y$. Proposition 1. ...
Jakobian's user avatar
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6 votes

Are separability and ccc equivalent for closed subspaces of $\beta N$?

No. There is a compactification of $\mathbb{N}$ whose remainder, $K$, is ccc non-separable. So there is a continuous surjection $f$ from $\beta\mathbb{N}\setminus\mathbb{N}$ onto $K$; take a closed ...
KP Hart's user avatar
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6 votes
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Is each Parovichenko compact space homeomorphic to the remainder of a soft compactification of $\mathbb N$?

Here is a partial answer: the Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko; the proof is a bit long, so I put it in a PDF-file on my website. Also, I retract my ...
KP Hart's user avatar
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6 votes
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Partitioning $\beta \mathbb{Z} \setminus \mathbb{Z}$

Yes, this is possible. First of all, let me suggest a way of thinking about the $+$ operation. If you're familiar with the idea of taking limits along an ultrafilter, then given $p,q \in \mathbb Z^*$,...
Will Brian's user avatar
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6 votes
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Best introductory texts on pointless topology

Topology via Logic - theoretical computer scientist Stone Spaces - pure mathematician Both are really good. Topology via logic as it gives a good account of domain theory, including power domains. ...
Christopher Townsend's user avatar
6 votes
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Embeddability into $\beta\omega$ and $\omega^*$

Answer to 1: In On closed subspaces of $\omega^*$ (Proc. AMS, 1993) it is shown by Dow, Frankiewicz and Zbierski that in the $\aleph_2$-Cohen model every compact zero-dimensional $F$-space of weight ...
KP Hart's user avatar
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6 votes

The Stone-Čech compactification of the fixed point set

Not without further assumptions. First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ ...
KP Hart's user avatar
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5 votes
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Borel $\sigma$-algebra in $\beta \mathbb N \times \beta \mathbb N$

The answer is no. Jiří Nedoma proved that if $(X,\Sigma)$ is a measurable space $|X| > 2^{\aleph_0}$, then the diagonal is not a measurable subset of $(X\times X, \Sigma \otimes \Sigma)$. (The ...
Robert Furber's user avatar
5 votes
Accepted

Tychonoff-ization and Urysohn (functionally Hausdorff) topological spaces

0a Correct, except for one point: is $X$ is not completely regular then there is no $\beta X$, so the third equivalence in `functionally Hausdorff' does not exist. 0b Not quite, the paper mentioned ...
KP Hart's user avatar
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5 votes

Is $\beta \mathbb{N}$ homeomorphic to its own square?

An indirect argument: Since the Banach space of continuous functions $C(\beta\mathbb{N})$ is isomorphic to $\ell_\infty$, it contains no complemented copies of $c_0$. Since $C(\beta\mathbb{N}\...
M.González's user avatar
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5 votes

Elementary equivalence between $n\mapsto n+1$ and its inverse on the Stone-Čech remainder?

Yes, these two structures are elementarily equivalent. This is proved as a corollary to another theorem, which states Theorem: CH implies that $\Phi$ and $\Phi^{-1}$ are conjugate to each other in the ...
Will Brian's user avatar
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