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2 votes

Which closed subsets $Y$ of a compact space $X$ admit a linear extensor $C(Y)\to C(X)$?

This is an extended comment expanding on the remarks I left in the comments. It's far from a definitive answer. To make the discussion less technical, I'll restrict to compact Hausdorff spaces so that ...
Tyrone's user avatar
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1 vote

What is the Lebesgue covering dimension of this topological space?

Concerning the Lebesgue covering dimension, absolutely nothing can be said, if you work with manifolds of total space-time dimension 3 or higher. Preamble We will consider spacetimes with closed ...
Willie Wong's user avatar
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1 vote
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Axiomatic definition of Katětov closure operator

There is a discussion of the relation between convergence spaces and closeness spaces in the paper S. Kasahara, Closeness spaces and convergence spaces, Proc. Japan Acad., 50(4) (1974), 303-308. The ...
Tyrone's user avatar
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1 vote

Notions of convergence not corresponding to topologies

The new notion of light condensed set, due to Dustin Clausen and Peter Scholze, has a notion of convergence, and indeed moving from the general setting of condensed sets to light condensed sets makes ...
David Roberts's user avatar
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0 votes

Axiomatic definition of Katětov closure operator

This is a partial answer. Let $X$ be a compact Hausdorff space and let $\mathcal{U}$ be an ultrafilter on a set $I$. Then any function $\phi: I \to X$ induces an ultrafilter on $X$, namely $\{A \...
Nik Weaver's user avatar
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2 votes

Is a Borel image of a Polish space analytic?

In this article by Repovš and Zdomskyy its shown that under MA$_{\omega_1}$, if $X$ is a regular space with countable tightness which contains an $L$-space, then $X^n$ is not Lindelof for some $n$. ...
Jakobian's user avatar
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4 votes
Accepted

Points in the Stone Cech compactification are intersection of open sets

Yes if the point is from $\mathbb{N}$ (it is isolated). No if the point is in $\beta\mathbb{N}\setminus\mathbb{N}$ because in that subspace every nonempty $G_\delta$-set has nonempty interior, see ...
KP Hart's user avatar
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3 votes
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Show that the distance converges to zero

$\newcommand\de\delta\newcommand\ep\varepsilon$You wrote Could you help me to show that under Ass1 and Ass2 $$d_H(A, A_n)\rightarrow_{a.s.} 0$$ Of course, this is not true in such generality. For ...
Iosif Pinelis's user avatar
14 votes
Accepted

What is this equivalence relation on topological spaces: there are bijective continuous maps in both directions

This relation was introduced (I don't know if for the first time) in the 1984 paper Bijectively related spaces I: Manifolds by P. H. Doyle and J. G. Hocking. As the title indicates, two spaces that ...
Ramiro de la Vega's user avatar
6 votes
Accepted

Does a coarser topology lead to a non-Hausdorff topological manifold?

Your hypothesis (modifying the topology to produce a non-Hausdorff topological manifold from a Hausdorff one) is impossible, if you assume the non-Hausdorff manifold has dimension equal or less than ...
Willie Wong's user avatar
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0 votes

Show that the distance converges to zero

EDIT: That whole answer assumes the $p_n$'s are uniformly and independently distributed on $[0,1]$. I have no idea on how to extend it to the general case of other distributions. It gets elementary ...
Claude Chaunier's user avatar
8 votes

On the classification of second-countable Stone spaces

The paper linked by YCor in the comments, R. S. Pierce. Existence and uniqueness theorems for extensions of zero-dimensional compact metric spaces, Trans. Amer. Math. Soc., 148:1–21, 1970, doi:10....
3 votes
Accepted

The separability of superextensions

There are non-separable zero-dimensional compact Hausdorff spaces with a separable superextension. Because of zero-dimensionality one obtains $\lambda X$ as the space of all maximal linked systems of ...
KP Hart's user avatar
  • 9,665
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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2 votes

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

The question is answered by Will Brian arXiv, Feb. 6 2024.
Mohammad Golshani's user avatar
8 votes
Accepted

Are Euclidean spaces $\Delta$-generated?

Yesterday, I answered in a comment, but I'll give a more detailed answer today so that this question doesn't linger on the unanswered queue. First, the idea of $\Delta$-generated spaces is just like &...
David White's user avatar
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5 votes

A continuous injection from the Hilbert cube to the real line?

There is not even a continuous injection $[0,1]^2 \to [0,1]$. If there were, we could compose it with an injection $[0,1] \to [0,1]^2$ taking $x$ to $(x,\frac{1}{2})$ to obtain a continuous injection $...
Andrej Bauer's user avatar
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3 votes

Is there a uniformly continuous injective image of $(0,1)\setminus\Bbb Q$ in the Cantor space?

Indeed it isn't possible. Consider any uniformly continuous image of those irrationals in the Cantor set. The inverse image $X$ of a clopen set, i.e. a finite union of the smaller "copies" ...
Pierre PC's user avatar
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2 votes
Accepted

Is the Mandelbrot set Suslinian?

There are different versions of the Suslinian property for continua. One is called finitely Suslinian, a term I think I learned about from the OP - if I recall correctly, it asks that, for any eps>...
Lasse Rempe's user avatar
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2 votes
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What is known about these "explicitly represented" spaces?

Your notion seems to be what would be called a $2^\omega$-based represented space following de Brecht, Schröder and Selivanov here (but there might be a misaligned in the details somewhere): https://...
Arno's user avatar
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9 votes
Accepted

Is every compact, sober, second-countable space the image of $2^\omega$?

We can built a counterexample by adding a bottom element to $\mathbb{N}^\mathbb{N}$. Let $\mathbb{N}^\mathbb{N}_\bot$ have the underlying set $\mathbb{N}^\mathbb{N} \cup \{\bot\}$, and let a set be ...
Arno's user avatar
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