39

There is no such topology.
Suppose there were. Then $\mathbb R$ itself is not compact, so there is an open cover $\mathcal U$ of $\mathbb R$ with no finite subcover. Using recursion, we can construct a non-compact countable subset of $\mathbb R$.
To begin, let $x_0 \in \mathbb R$ and let $U_0$ be any member of $\mathcal U$ containing $x_0$. Since $\{U_0\}$ ...

16

The answer is yes. In the paper "Compactness for omitting of types" (Annals of Mathematical Logic, vol 14(1) (1978), pp. 39-56), Benda gives such a characterization. He proves the following:
Theorem. The following are equivalent:
1) $\kappa$ is supercompact,
2) If $T$ is a theory in $L_{\kappa, \kappa}$ and $\Sigma(x,y)$ is a type such that $\{ \Delta \in ...

answered Mar 29 '14 at 4:53

Mohammad Golshani

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14

To me, the second definition of local compactness is much to be preferred for the simple reason that such locally compact spaces $X$ are exponentiable in $Top$, meaning that $X \times -: Top \to Top$ has a right adjoint $(-)^X: Top \to Top$ (even without the Hausdorff condition), and all this implies (such as $X \times -$ preserving coequalizers). In fact ...

answered Feb 25 '17 at 4:14

12

Proposition. Let $X$ be a topological space. Then $X$ is a compact object if and only if $X$ is a finite discrete space.
Before giving the proof, we state an easy lemma.
Lemma. Suppose $X$ is a compact object in $\operatorname{\underline{Top}}$. Then $X$ is finite.
Proof. Let $Y$ be the indiscrete topological space with underlying set $X$; i.e. $\...

answered Dec 16 '17 at 23:13

R. van Dobben de Bruyn

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11

This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, \frac{1}{2n})\cup [\frac{2}{2n}, \frac{3}{2n}] \cup\dots\cup [\frac{2n-2}{2n}, \frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $\chi_{A_n}$ converges to the function $\frac{1}{2}$ in the weak* topology. ...

11

No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X = \ell_\infty^*$. Then you have no chance of building such operators.
Another possibility for counterexamples comes with non-separable reflexive spaces (or, ...

10

I happened to run across an answer to 2. (and therefore 1.) just by clicking on one of the related MO links, finding the same question in a comment by Timothy Gowers under MO16578 that was later answered in a comment by Anton Petrunin, and then googling for more information. The key term you want seems to be "pseudo-arc", apparently a construction very ...

answered May 24 '14 at 18:22

8

In ${\mathbb R}^3$, try three solid cones with different heights and the same base radius, axis and vertex (the vertex being the intersection of the three boundaries).

8

I feel like I should add that there exists a nice characterization due to Tony Wickstead of ordered Banach spaces whose compact sets are order bounded.
Wickstead A.W., Compact subsets of partially ordered Banach spaces, Mathematische Annalen (1975) (MSN)

7

One way of disproving this conjecture is to construct a norm convergent sequence in E which is not order bounded.
E.g. the sequence $(\frac{1}{n}e_n)$ in $l^1$.

7

A space is anti-compact iff it has no proper covers consisting of two sets, or equivalently if the intersection of any two nonempty closed sets is nonempty. This is equivalent to the specialization order being directed downwards.
We can use this to prove any anti-compact space is anti-metacompact, so your three conditions are equivalent. Suppose $X$ is ...

7

Any LindelĂ¶f non-metrizable Hausdorff space will do (EDIT: you need that the space is C-closed as well, see below), but more generally, a space is called isocompact iff every closed countably compact subset of X is compact, cl-isocompact iff the closure of a countably compact subset is compact, and C-closed iff any countably compact subset is closed. This ...

7

In addition to the above examples, it might be of interest that functional analysis is awash with such spaces. For example, the space of distributions on a compact interval or that of functions analytic on a closed domain in the complex plane, both with their natural topologies. More generally, any Silva space has this property. A further ubiquitous ...

6

Take the classical Bernoulli scheme with the base $(1/2,1/2)$, and let $A_n$ be the set where the $n$-th coordinate is 1. Then $d(A_n,A_m)=1/2$ whenever $n\neq m$. Since all purely non-atomic Lebesgue spaces are isomorphic, the same is applicable to any purely-non-atomic Borel measure on a Polish space.

6

More is true: if K is compact on $L^{p_1}$ for some $1\leq p_1<\infty$ and bounded on $L^{p_2}$ for some other $1 \leq p_2 \leq \infty$, then it is compact for all $p$ in the interval $[p_1,p_2)$ or $(p_2,p_1]$. This is a results of KrasnoselĘąskiÄ. See http://www.ams.org/mathscinet-getitem?mr=119086
Interestingly it is an open question whether this holds ...

answered Feb 6 '15 at 8:15

Mikael de la Salle

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6

You can find in many functional analysis text books the theorem that the closed convex hull of a weakly compact subset of a Banach space is weakly compact. But what you want is simpler than the general theorem. Here is a simple conceptual proof: Let $(y_n)$ be a weakly null sequence in $X$ and consider the bounded linear operator $T:\ell_1 \to X$ that maps ...

6

You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.
Remark. Note that $\...

answered Jan 20 '16 at 3:04

R. van Dobben de Bruyn

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6

Are you asking if there is a $c>1$ and a measure $\nu$ with this property? And is the first condition supposed to hold with a uniform upper bound? (If not, any probability measure satisfying the second condition would work, as Steve points out.)
If so, you are asking for the existence of a non-trivial "doubling measure". A necessary and sufficient ...

6

Yes it's true. It essentially follows from a lemma quoted in the linked answer by user Qayum Khan, which I quote verbatim:
Neighboring subgroups theorem [1942]
Any compact subgroup $H$ of an arbitrary Lie group $G$ admits a neighborhood $O$ in $G$ such that any subgroup contained in $O$ is $G$-conjugate to a subgroup of $H$.
Now let by contradiction $(...

5

Yes: think of three closed balls in $\mathbb{R}^3$, such that the intersection of the boundaries of the first two balls is a small circle tangent to the boundary of the third ball from inside.

5

One of my favorite spaces has this property: the extended long ray.
First, the long ray itself: this is just the space $L$ gotten by pasting together $\omega_1$-many copies of $[0, 1)$ in the natural way. Formally, $L$ is the lexicographic order on $\omega_1\times[0, 1)$, with both viewed as linear spaces in the natural way.
Now, $L$ is densely ordered, ...

answered Jun 23 '14 at 18:17

Noah Schweber

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5

Rauniâ€™s link to the Freeman et al paper perhaps provides sufficient excuse for me to give here a simple proof (which, however, uses some theory) of the theorem in that paper. The background can more or less be found in Diestelâ€™s book â€śSequences and Series in Banach Spacesâ€ť, Springer GTM 92. The theorem says that if every weakly compact subset of the Banach ...

5

I'll start with several known facts.
Proposition 1. Suppose that $E$ is a real Hilbert space with norm $\Vert -\Vert$ and $K: E\to E$ is a a compact, selfadjoint
positive operator. Denote by $R(K)$ the range of $K$. Equip the
range with the norm
$$ \Vert x\Vert_K:=\Vert K^{-1} x\Vert,\;\;x\in R(K). $$
Then the inclusion $(R(K), \...

answered Dec 15 '15 at 11:09

Liviu Nicolaescu

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5

I think any compact, first countable, non-metrizable Hausdorff space suffices, since then any countably compact set is closed and hence compact. In Steen and Seebach's Counterexamples in Topology, #95, the weak parallel line topology, has these properties.

5

Regularity is enough. Given an open cover $\mathcal{A}$ of $X$ we can use regularity to build another cover $\mathcal{B}$ such that for any $U \in \mathcal{B}$ there is a $V \in \mathcal{A}$ with $\overline U \subseteq V$. Now a finite subset of $\mathcal{B}$ covering $D$ naturally provides a finite subset of $\mathcal{A}$ that covers $X$.
Completely ...

answered Dec 31 '17 at 16:44

Ramiro de la Vega

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5

Of course you can, and this is how Arzela-Ascoli is often proved. You may fix a finite $\varepsilon/3$-net $D\subset X$ and partition $[0,1]$ onto disjoint subsets $A_1,\ldots,A_N$ of diameter less than $\varepsilon/3$. For any 1-Lipschitz function $f:X\rightarrow [0,1]$ we consider the function $[f]:D\rightarrow \{1,2,\ldots,N\}$ defined as $[f](t)=i$ iff $...

4

Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact, $Y \subseteq X$, and every open cover of $X$ has a finite subcover of $Y$. Consider the covering of $X$ by all open subsets of whose closure is compact. Since finitely ...

4

This is an answer to your question 1):
Yes: every compact complex manifolds of dimension one (also known as a Riemann surface) admits a nonconstant meromorphic function. Even more is true: for every point $p$ in your Riemann surface, there exists a non-constant meromorphic function that is regular away from the point $p$.
I do not understand your question ...

answered Jun 5 '14 at 23:34

André Henriques

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4

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.
If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then ...

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