Questions tagged [compactness]

The tag has no usage guidance.

Filter by
Sorted by
Tagged with
44 votes
17 answers
16k views

What is your favorite proof of Tychonoff's Theorem?

Here is mine. It's taken from page 11 of "An Introduction To Abstract Harmonic Analysis", 1953, by Loomis: https://archive.org/details/introductiontoab031610mbp https://ia800309.us.archive....
31 votes
3 answers
5k views

Is "compact implies sequentially compact" consistent with ZF?

Over at the nForum, we've been discussing sequential compactness. The discussion led me to realise that I naively assumed that nets were simply Big Sequences, and that I could make a reasonable guess ...
Andrew Stacey's user avatar
27 votes
1 answer
832 views

Can closed compacts in a topological group behave "paradoxically" with respect to unions, intersections, and one-sided translations?

Consider two closed compacts $A$ and $B$ in a topological group $\Gamma$. Let $A'$ be a left translation of $A$ and $B'$ a left translation of $B$: $A' = aA$, $B' = bB$. Suppose it is known that $A'\...
Alexey Muranov's user avatar
17 votes
5 answers
822 views

How can one characterise compactness-by-experiment?

There are a myriad different variations on the theme of "compactness", and some of them have even made it on to Wikipedia. I'm interested in finding out more about types of compactness that ...
Andrew Stacey's user avatar
15 votes
2 answers
883 views

Does there exist a supercompactness theorem?

Large cardinals such as weakly compact cardinals, measurable cardinals, strongly compact cardinals, and extendible cardinals all can be characterized in terms of a certain compactness theorem of ...
Joseph Van Name's user avatar
13 votes
1 answer
1k views

A topology on $\Bbb R$ where the compact sets are precisely the countable sets

QUESTION. In there a topology on $\Bbb R$ where the compact subsets are precisely the countable subsets? I am trying to create a counterexample to a certain claim, and I found that what I need is a ...
Cauchy's user avatar
  • 233
13 votes
1 answer
580 views

A generalization of the Arhangelskii Theorem

Arhangeleskii's Theorem states the following For any Hausdorff topological space $X$, $$ |X|\leq2^{\chi(X)L(X)} $$ where $\chi(X)$ is the character of $X$ and $L(X)$ is the Lindelöf degree of $...
user avatar
11 votes
1 answer
2k views

What are compact objects in the category of topological spaces?

Let $\mathscr C$ be a locally small category that has filtered colimits. Then an object $X$ in $\mathscr C$ is compact if $\operatorname{Hom}(X,-)$ commutes with filtered colimits. On the other hand, ...
R. van Dobben de Bruyn's user avatar
11 votes
2 answers
437 views

When is a k-space locally compact?

We're looking at the possible cardinal sequences of LCS (locally compact, Hausdorff, scattered) spaces, which has led us to think about taking a quotient of a locally compact, scattered space. A k-...
Carla Simons's user avatar
11 votes
1 answer
276 views

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

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right ...
Pietro Majer's user avatar
  • 56.5k
11 votes
0 answers
255 views

A ZFC-example of a countably compact paratopological group which is not a topological group

Problem. Does there exist a ZFC-example of a countably compact Hausdorff paratopological group which is not a topological group? (The problem posed 27 May 2015 by Alexander Ravsky on page 9 of Volume ...
Lviv Scottish Book's user avatar
10 votes
0 answers
182 views

Masas in SAW*-algebras

I asked this question three years ago at MSe but it has no response; let me try here. Pedersen distilled the following class of C*-algebras which he termed SAW*-algebras (Journal of Operator Theory, ...
Tomasz Kania's user avatar
  • 11.3k
9 votes
3 answers
750 views

Structure theorems for compact sets of rationals

Everyone knows the Heine-Borel theorem characterizing compact subsets of Euclidean space. For any $n \in \mathbb N$ a set $A \subseteq \mathbb R^n$ is compact just in case it is closed and bounded (in ...
Corey Bacal Switzer's user avatar
9 votes
2 answers
771 views

On the definition of locally compact for non-Hausdorff spaces

It seems that there are different conventions in the literature as to what is a locally compact space (when the space is not supposed Hausdorff). The two main non equivalent definitions I've seen ...
Phil-W's user avatar
  • 975
9 votes
1 answer
406 views

Do compactly generated spaces have a more direct definition?

Is there an elementary way to define Haussdorf-compactly generated weakly Hausdorff topological spaces in a way that does not need defining topological space first? Weakly Hausdorff sequential spaces ...
saolof's user avatar
  • 1,823
8 votes
3 answers
587 views

Is there a non-metrizable topological space for which any countably compact subset is compact?

The title is the question : Is there a non-metrizable topological space for which any countably compact subset is compact ? EDIT : non-metrizable and Hausdorff
Michael's user avatar
  • 361
8 votes
1 answer
258 views

Characterization of pretty compact spaces

This is a cross post from MSE. I believe that the following problem have already been considered by some sophisticated topologist. Definition 1. A non-compact Hausdorff topological space $X$ is called ...
Norbert's user avatar
  • 1,687
8 votes
2 answers
559 views

Totally disconnected subspaces

This question is motivated by this one, where no simple solution within ZFC seems to exist. Let me ask a weaker question then. Suppose that $K$ is a compact, Hausdorff, non-metrizable space. Does it ...
spooky's user avatar
  • 81
8 votes
1 answer
192 views

Subspaces of $L_p([0,1])$ whose unit ball is compact for the topology of convergence in measure

Any information about the following questions would be welcome. I wonder whether there are (well-known or easy) closed and infinite dimensional subspaces of $L_p([0,1])$ ($1<p<\infty$) whose ...
user159631's user avatar
8 votes
1 answer
1k views

Noncompactness of the Sobolev embedding in the critical exponent case

Let $\Omega \subset \mathbb R^n$ be a bounded domain with a Lipschitz boundary and $n > p \ge 1$. It is well known that up to the critical exponent $p^* = pn/(n − p)$, i.e. $q < p^*$, the ...
anonymous's user avatar
  • 436
7 votes
1 answer
785 views

Compactness of set of indicator functions

Let $\chi_A(x)$ denote an indicator function on $A\subset [0,1]$. Consider the set $$K=\{\chi_A(x): \text{ A is Lebesgue measurable in }[0,1]\}.$$ Is this set compact in $L^\infty(0,1)$ with respect ...
Saj_Eda's user avatar
  • 395
7 votes
1 answer
846 views

Is a closed subset of an extremally disconnected set again extremally disconnected?

Let $T$ be a compact Hausdorff extremally disconnected set (so $T$ is a compact Hausdorff space, such that the closure of each open subset is again open). Let $S \subseteq T$ be a closed subset. ...
AlexIvanov's user avatar
7 votes
2 answers
402 views

Compact operators on Lebesgue spaces

Let $K:{\rm L}^p({\bf R}^d)\to {\rm L}^p({\bf R}^d)$ be a bounded linear operator for every $p\in(1,\infty)$. Assume that for some $r\in(2, \infty)$ it holds that $K$ is compact on ${\rm L}^q({\bf R}^...
Semmel's user avatar
  • 165
7 votes
1 answer
316 views

Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?

Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$. Of course any ...
Dominic Else's user avatar
7 votes
2 answers
478 views

Is the union of a compact and the relatively compact components of its complementary in a manifold compact?

I was thinking of a way to prove this and I realised that for my approach the lemma from the title would be useful, and it´s an interesting question on its own. Obviously it is true if the manifold is ...
Saúl RM's user avatar
  • 7,916
7 votes
2 answers
967 views

CG spaces from the perspective of sheaves over compact Hausdorff spaces

A compactly generated space is a space $X$ such that $f : X \rightarrow Y$ is continuous if and only if $K \rightarrow X \stackrel{f}{\rightarrow} Y$ is continuous for each compact hausdorff space $K$....
Ronald J. Zallman's user avatar
7 votes
1 answer
720 views

Fréchet-Kolmogorov compactness Theorem for Lp spaces on manifolds

Suppose I have a family of functions $\mathcal{F} \subseteq L^2(\mathcal{M}, P)$ where $\mathcal{M}$ is a compact manifold, and $P$ is a probability distribution on $\mathcal{M}$. Is there an ...
Barrett's user avatar
  • 143
7 votes
1 answer
785 views

Weak*-convergence of signed measures

Let $X$ be a compact Hausdorff space and let $M(X)$ denote the space of signed measures that is naturally dual to $C(X)$, the space of continuous functions on $X$. I am interested whether the ...
user3522356's user avatar
6 votes
14 answers
5k views

Applications of compactness [closed]

Similar to this question: Applications of connectedness I want to collect applications of compactness. E.g.: compact + discrete => finite, which can be used to prove the finiteness of the ...
6 votes
2 answers
778 views

Is every continuous action of a compact topological group closed?

I am reading Bredon's Introduction to compact transformation groups, and came across the following result and proof on page 34: Even though he writes "Recall our standing assumption that $X$ is ...
Ben's user avatar
  • 1,010
6 votes
3 answers
765 views

Convolution of $L^2$ functions

Let $u\in L^2(\mathbb R^n)$: then $u\ast u$ is a bounded continuous function. Let me assume now that $u\ast u$ is compactly supported. Is there anything relevant that could be said on the support of $...
Bazin's user avatar
  • 15.1k
6 votes
2 answers
2k views

Is there an easier proof to show that the closed convex hull of a normalized weakly null sequence is weakly compact?

In a paper that I am reading there is a following step: Let $X$ be a Banach space and let $(x_k) \subset X$ be a normalized sequence that converges weakly to $0$. Then $\overline{co}(x_k)$ is a ...
Rauni's user avatar
  • 163
6 votes
2 answers
265 views

Is every compact simply-connected reductive p-adic group perfect?

Let $k$ be a nonarchimedean local field and $G$ a reductive $k$-group, which we assume to be semisimple and simply-connected. Recall that an abstract group $H$ is perfect if it is generated by ...
David Schwein's user avatar
6 votes
2 answers
2k views

How do I prove that compact-open topology is metrizable?

Let $X$ be a $\sigma$-compact topological space and $(Y,d)$ be a metric space. Let $\{K_n\}$ be a sequence of compact subsets of $X$ whose union is $X$. Define $\rho_n(f,g):=\sup \{d(f(z),g(z)): z\...
Rubertos's user avatar
  • 337
6 votes
2 answers
988 views

Is the separability of the space needed in the proof of the Prohorov's theorem?

The Section 5 of the book: Billingsley, P., Convergence of Probability Measures, 1999, studies Prohorov's theorem. A short reminder is given below. Let $\Pi$ be a family of probability measures on ...
Mark's user avatar
  • 647
6 votes
1 answer
257 views

Topologies that turn the real numbers into a compact Hausdorff topological group

If I'm not mistaken, the question I put on the title used to be on this site, but I'm not being able to find it at all. I'm therefore reposting it so that someone can either give me the old link or ...
Pedro Lourenço's user avatar
6 votes
1 answer
846 views

Rellich-Kondrachov compacteness Theorem for the Euclidean space with Gaussian measure

Let $\gamma_n: \mathbb{R}^n\to\mathbb{R}$ be the Gaussian distribution function defined by $$ \gamma_n(x):=(2 \pi)^{-\frac{n}{2}} e^{-\frac{|x|^2}{2}}. $$ Let $d\gamma_n$ denote the following measure ...
Lorenzo Cavallina's user avatar
6 votes
1 answer
199 views

When does base-change in topological spaces preserve quotient maps?

The question when $(-) \times X$ preserves colimits in topological spaces is well-studied. Since it always preserves arbitrary coproducts (disjoint unions), one only has to show when it preserves ...
Lennart Meier's user avatar
6 votes
1 answer
539 views

When Stone–Čech compactification is totally disconnected

A topological space $X$ is totally disconnected if the connected components in $X$ are the one-point sets, and a topological space, $X$ is called completely regular exactly in case points can be ...
Arena's user avatar
  • 61
6 votes
1 answer
576 views

Uniqueness of limits and compactness implies closure

It is not difficult to prove that in a Hausdorff topological space every compact set is closed, and almost trivial that if in a topological space X every compact set is closed then X is T1. As ...
Daniel Elessar's user avatar
6 votes
1 answer
288 views

Is there a compactification with nontrivial connected remainder?

Question: Let $X$ be a continuum and $p \in X$. Under what conditions does there exist a compactification $\gamma (X-p)$ with $\gamma (X-p) - (X-p)$ connected and nondegenerate? Throughout, $X$ is a ...
Daron's user avatar
  • 1,761
6 votes
0 answers
189 views

Making the analogy of finiteness and compactness precise

If one asks about the intution behind compact topological spaces, most often one will hear the mantra “Compactness of a topological space is a generalisation of the finiteness of a set.” For example,...
Jannik Pitt's user avatar
  • 1,163
5 votes
1 answer
463 views

Who are the owners of the compactness theorem in $L^p(\Bbb R^d)$?

As the title says, I am interested to know Who are the owners of the compactness theorem in $L^p(\Bbb R^d)$. There is some confusion in the literature. Let recall that the compactness theorem in $L^p(\...
Guy Fsone's user avatar
  • 1,033
5 votes
2 answers
617 views

Beyond Cantor's Teepee

From Counterexamples in Topology by Steen and Seebach (2nd edition) example 129 page 145 we have an example of connected and totally path-disconnected space. It is defined as follow: Fix $p= (1/2,1/2)...
Portland's user avatar
  • 2,752
5 votes
1 answer
802 views

Existence of injective compact operators

We know that if $X$ is a separable Banach space, then for every infinite dimensional Banach space $Y$, there exists an injective compact operator from $X$ to $Y$. My query is for every Banach ...
Anupam's user avatar
  • 479
5 votes
2 answers
571 views

Anti-compactness

Let $(X,\tau)$ be a topological space such that $\tau\ne\{\emptyset\ X\}.\ $ We call an open cover $\mathcal{U}$ of $(X,\tau)$ proper if $\ X\notin \mathcal{U}.\ $ Moreover we say that $(X,\...
Dominic van der Zypen's user avatar
5 votes
1 answer
243 views

Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of ...
saolof's user avatar
  • 1,823
5 votes
1 answer
354 views

Mapping scheme from a proper variety

Let $X$ be a proper scheme over a field $k$. Let $T$ be a scheme over $k$. Is it true that morphisms $T \times X \to \mathbb{A}^1$ are in bijection with morphisms $T \to \Gamma (X, \mathcal{O}_X)$ (...
Sasha's user avatar
  • 5,492
5 votes
2 answers
171 views

Polish space isometric to its hyperspace

For a Polish space $(X,d)$ its hyperspace $(K(X),d_H)$ is also a Polish space. (Here $K(X)$ denotes the set of all nonempty compact subsets of $X$, and the Hausdorff metric $d_H$ is defined by $d_H(K,...
chj's user avatar
  • 157
5 votes
2 answers
805 views

Covering compactness in the weak sequential topology

Let $X$ be a real Banach space. Apart from the norm topology, we can consider the following weak topologies on $X$: the weak toplogy, defined as the initial topology with respect to $X^*$. In other ...
Daniel Steck's user avatar