39
votes
Accepted
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 ...
- 18.5k
23
votes
Accepted
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 ...
- 18.5k
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 ...
- 41.8k
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 ...
- 18.5k
16
votes
Accepted
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 ...
- 31.2k
14
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\...
- 18.4k
11
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\...
- 63.9k
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 ...
- 73.1k
10
votes
Accepted
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 ...
- 11.4k
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 ...
- 1,832
10
votes
Accepted
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 $\...
- 38.6k
9
votes
Accepted
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 $\beta\omega-\omega$ ...
- 73.1k
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\...
- 11.4k
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 ...
- 41.8k
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 ...
- 32.5k
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,\...
- 11.4k
7
votes
Accepted
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.
...
7
votes
Accepted
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: ...
- 11.4k
7
votes
Accepted
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. ...
- 1,201
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 ...
- 11.4k
6
votes
Accepted
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 ...
- 11.4k
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, $\Omega$ ...
- 11.4k
6
votes
Accepted
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 $p,q \in \mathbb Z^*$,...
- 18.5k
6
votes
Accepted
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 ...
- 11.4k
6
votes
Accepted
Stone-Čech boundary is not extremally disconnected
We can suppose $X=\omega$. Let $(X_i)_{i\in I}$ be a continuum family of infinite subsets of $\omega$ with pairwise finite intersection. Define $Y_i=\bar{X_i}-X_i$. So the $Y_i$ are pairwise disjoint ...
- 63.9k
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 ...
- 11.4k
5
votes
Accepted
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 ...
- 3,768
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}\...
- 4,461
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 ...
- 18.5k
Only top scored, non community-wiki answers of a minimum length are eligible