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20 votes

Why are extremally disconnected spaces so hard to give examples of?

We can always get non-principal ultrafilters on sets from non-discrete extremally disconnected spaces. Let $X$ be a topological space. Let $\text{Clo}(X)$ denote the Boolean algebra of clopen subsets ...
Joseph Van Name's user avatar
12 votes
Accepted

Topos notions coming from topology and uniqueness of generalizations

If the absence of adjoints is what worries you, you can consider this to be a two-step process - and I would argue that in practice this is the case in the vast majority of cases: One first generalize ...
Simon Henry's user avatar
  • 41.3k
9 votes

On a metrized $n$-dimensional manifold $X$, does every $x \in X$ have a small ball $B_\delta(x)$ that is homeomorphic to $\mathbb R^n$?

Consider $f:\mathbb R\to\mathbb R$, $f(x)=x\sin(1/x)$ with $f(0)=0$. Then $$d(x,y)=|x-y|+|f(x)-f(y)|$$ defines a metric on $\mathbb R$ which induces the usual topology (because of the continuity of $f$...
Jochen Wengenroth's user avatar
6 votes

How to properly define a slice knot (or a locally flat disk)?

The question is, what do you want to achieve by local flatness? One possible answer is the following: You want your slice disk to have a tubular neighborhood. To make sure that you have a tubular ...
Stefan Friedl's user avatar
6 votes
Accepted

On connected sum of compact manifolds along a submanifold

Note that "closed" means "compact without boundary". Here are answers. (1): Yes, if $M_1$ and $M_2$ are closed manifolds then $M_1\sharp_KM_2$ is a closed manifold. If $M_1$ and $...
Gael Meigniez's user avatar
6 votes
Accepted

Perfectly normal but not collectionwise normal space in ZFC

Bing's Example H is found in R. Bing, Metrization of Topological Spaces, Canadian J. Math. 3 (1951), 175-186. It is on the page following his more famous Example G. It is constructed in ZFC and has ...
Tyrone's user avatar
  • 5,326
6 votes
Accepted

Can a scattered profinite set continuously surject onto a non-scattered profinite set?

No this is not possible. There might be an easier proof than this but I don't know one off the top of my head. First note that a profinite set is scattered if and only if no non-empty closed subspace ...
James E Hanson's user avatar
5 votes

Category of topological spaces with open or closed maps

A nice recent paper of Bezhanishvili and Kornell (which actually references this MathOverflow question specifically) has shown that the existence of products fails very badly in $\mathrm{Top}_{\mathrm{...
James E Hanson's user avatar
5 votes

How algebraic can the dual of a topological category be?

I have not properly digested your framework of definitions; the following example may or may not fit within it. (If not, it would be helpful to explain why not.) Let $\operatorname{CH}$ be the ...
Neil Strickland's user avatar
4 votes

Menger and Scheepers subsets of $\mathbb R$

I believe that the sets in the paper that you cite (subsets of the Cantor space that contain the set of eventually 0 elements) are homeomorphic to sets of reals that contain the rational numbers, by a ...
Boaz Tsaban's user avatar
  • 3,084
4 votes

A finite dimensional continuum with a subset $A$ such that both $A$ and $X\setminus A$ are dense and contractible

A fancy decomposition of the closed unit disk $X$ of $ \mathbb{C}$ into two dense contractible sets. Let $D_1$, $D_2$ be a $2$-partition of the interval $[0,1]$ into dense sets. Consider the ...
Pietro Majer's user avatar
  • 58.9k
4 votes

A question on unit norm elements of $\ell^2 \setminus \bigcup_{0<p<2 }\ell^p$

Umm, $A$ is path-connected and it is fairly simple -- just change the coordinates one by one. Say we want to go from $u\in A$ to $v\in A$. On $[0, \frac{1}{2}]$ change linearly $u_1$ to $v_1$, then on ...
Aleksei Kulikov's user avatar
4 votes
Accepted

Stone-Čech compactification of a Boolean subalgebra of $\{0,1\}^S$

A partial answer. for your specific example, the algebra $B$ generated by the rational intervals, the answer is negative. This algebra is countable and dense in $2^\mathbb{R}$. It is countable and ...
KP Hart's user avatar
  • 10.8k
4 votes
Accepted

Is any submetrizable linear topology linearly submetrizable?

Yes. Let $(X,\tau)$ be a toplogical vector space and $d$ a metric on $X$ generatin a coarser topology $\pi$. Recursively, one finds balanced $\tau$-neighbourhoods $U_n$ of the origin with $$ U_{n+1}+...
Jochen Wengenroth's user avatar
4 votes

Can a scattered profinite set continuously surject onto a non-scattered profinite set?

Here's the Stone duality perspective: Scattered profinite sets (a.k.a. scattered Stone spaces) correspond under Stone duality to superatomic Boolean algebras: $B$ is superatomic if every quotient ...
Alex Kruckman's user avatar
3 votes

How algebraic can the dual of a topological category be?

Check out the following papers. Barr and Pedicchio, $\text{Top}^\circ$ is a quasi-variety Barr and Pedicchio. Topological spaces and quasi-varieties. Dimov, Pedicchio, and Tironi. Frames and Grids. ...
Ivan Di Liberti's user avatar
3 votes

On a metrized $n$-dimensional manifold $X$, does every $x \in X$ have a small ball $B_\delta(x)$ that is homeomorphic to $\mathbb R^n$?

Regarding the first question: if the metric is the Carnot-Caratheodory metric on step-two sub-Riemannian manifolds, small balls should be homeomorphic to standard balls. (It is known that Carnot-...
A. Lerario's user avatar
3 votes
Accepted

Continuous collections of arcs

Let $F(c)=f^{-1}(c)$, $F: C\to K(X)$, where $K(X)$ is the hyperspace of $X$. Then $F$ is continuous by your condition and thus $F(C)$ is compact. Consequently $\bigcup F(C)$ is compact as well and $\...
Benjamin Vejnar's user avatar
2 votes

Universally closed implies proper for locales

It turns out that Vermeulen has essentially answered the question in [A note on stably closed maps of locales]. The argument there implies: Theorem. Let $g : X \to S$ be a morphism of locales. The ...
Zhen Lin's user avatar
  • 15.1k
2 votes

A question on unit norm elements of $\ell^2 \setminus \bigcup_{0<p<2 }\ell^p$

For the first question, yes, $S \setminus A$ is contractible. Rewrite the standard basis so that it is indexed by $\mathbb{Z}$. Let $U$ be the bilateral shift. Then $U$ is an isomorphism on $\ell^p$ ...
David Gao's user avatar
  • 2,344
2 votes

Stone-Čech boundary is not extremally disconnected

The question has already been answered satisfactorily, but I think the following presentation, which is implicit in the second part of YCor's answer, will clarify things. The fact that $\beta\mathbb{...
Gro-Tsen's user avatar
  • 31.9k
2 votes

Does regular $G_\delta$ imply normal?

Edit: I made a mistake in disagreeing with Henno, who after all is right. But perhaps someone can learn from my mistake. It is worth noting that a space in which every closed set is "regular $G_\...
Nick Mendler's user avatar

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