27
votes
Is there a general theory of "compactification"?
There's a distinction that I find striking but don't know how to formalize usefully or how to evaluate its importance: In algebraic geometry, moduli spaces get compactified, and this involves adding a ...
18
votes
Accepted
Is the one-point compactification of $\mathbb{N}$ computably countable?
The answer is no. Suppose that there are computable functions $q$ and $s$ as you describe.
Let $k$ be a program that performs the following task. It starts enumerating $1$s at the start of the ...
14
votes
Accepted
Do all homogeneous spaces have homogeneous compactifications?
Since you want a connected example:
A surface of infinite genus has no homogeneous compactification.
Indeed first observe a dense locally compact subset has to be open.
So the surface has to be open, ...
12
votes
Accepted
One point compactification of $(\mathbb{C}^{\ast})^n$
If $\widehat X$ is the 1-point compactification of $X$, then there is a homeomorphism (for, say, locally compact Hausdorff spaces)
$$
\widehat{X \times Y} \cong \widehat X \wedge \widehat Y
$$
with ...
12
votes
Do all homogeneous spaces have homogeneous compactifications?
The countable discrete space $\omega$ is a counterexample.
Suppose $Y$ is a homogeneous compactification of $\omega$, with $X \subset Y$ being homeomorphic to $\omega$. As $Y$ is infinite, it ...
10
votes
Accepted
Is each compactification of $\mathbb N$ soft?
Let $A=\{0,2,4,\dots\}$ be the even numbers and let $B=\{1,3,5,\dots\}$ be the odd numbers. Topologize $A\cup \beta B$ so that $A$ is a sequence limiting to a unique point in $\beta B \setminus B $. ...
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 ...
9
votes
Accepted
Uniqueness of smooth compactification upto a smooth morphism
This desire fails on a much more basic level. A smooth morphism is flat, and you can't even have a flat morphism between two smooth compactifications.
A flat morphism has equidimensional fibers. If ...
8
votes
Accepted
Sigma algebras on the Stone–Čech compactification of a countable discrete group
No.
First of all, take note that for compact zero-dimensional spaces $X$, the $\sigma$-algebra generated by all clopen sets is precisely the $\sigma$-algebra of Baire sets (recall that in a ...
8
votes
Accepted
Possible cardinalities of the remainders of compactifications of $\Bbb R$
Every connected compact Hausdorff space of weight $\aleph_1$ is the remainder $v \mathbb R \setminus \mathbb R$ of some compactification of $\mathbb R$. In particular, $[0,1]^{\aleph_1}$ is the ...
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,\...
7
votes
Accepted
Compactification of 6d (2, 0) SCFT on 4-manifolds
You could have a look at https://arxiv.org/abs/1806.02470 and references therein.
EDIT (taking into account the comment): compactification of the $\mathcal{N}=(2,0)$ 6d superconformal field theory on ...
6
votes
Accepted
Wonderful compactification
I am just writing my comment above as an answer, just in the special case that the parabolic is a Borel subgroup $B$. In this case, the reductive part $M$ is a maximal torus $T$ in $G$. Denote by $W\...
Community wiki
6
votes
Accepted
Minimal Nagata-like compactification
The following argument shows that the very best thing you could hope for does not work. I'm not sure whether I need all of my assumptions.
Claim: There does not exist a compactification ...
6
votes
Compactification theorem for differentiable manifolds ?
I'm just trying to give a partial solution to the second question.
Those examples had been mentioned in the previous posts. That is, finding manifolds which are open interiors of compact manifolds ...
6
votes
Which points in the Samuel compactification of a metric space $X$ are limits of uniformly discrete subsets of $X$?
A metric space $X$ is called isometrically homogeneous if for any points $x,y\in X$ there exists a bijective isometry $f:X\to X$ such that $f(x)=y$.
For isomemtrically homogeneous spaces this ...
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 ...
5
votes
Accepted
Naive compactification of $\mathbb{C}^*$-fibrations
No that is not true. There is a much simpler example than in the MO answer above. I might have written the following example in one of my earlier MO answers.
Let the proper target of the morphism ...
Community wiki
5
votes
Is the conformal compactification of $M \setminus \{ p \}$ unique?
Theorem 1.4 in C. Frances' preprint "Rigidity at the boundary for conformal structures and other Cartan geometries" asserts that the geodesic compactification is unique (up to conformal diffeomorphism)...
5
votes
Sheaf (Gieseker) compactification of moduli space of vector bundles
This is perhaps more of a comment, but it's too long.
I'm coming from the algebraic side of things, but it seems to me that there are several issues here. Be warned that when I say curve I mean a (...
5
votes
Nowhere compact subsets of the plane
According to the definition here a topological space is nowhere compact if every compact subset has empty interior. Assuming that's that you mean by "nowhere compact", the following set $X$ seems to ...
5
votes
Compactification of a product of manifolds
Yes, the quotient $C_M = \overline{M} \times \overline{\mathbb{R}} / \left\{\{x\} \times \overline{\mathbb{R}} : x \in \partial\overline{M}\right\}$ seems to do the job, where I mean that the points ...
4
votes
Accepted
Bohr compactification as a topological compactification
As Francois Ziegler answered, for a locally compact group $G$, the Bohr compactification of $G$ is a compactification in the usual sense iff $G$ is compact. This is true with no further restrictions.
...
4
votes
Bohr compactification as a topological compactification
The answer is no: unless $G$ is compact, its subspace topology inside $bG$ (called its Bohr topology) is much weaker than its own. So $\iota:G\to bG$, while continuous with dense image, is not an ...
4
votes
Stone-Cech compactification of $\mathbb{R}^n$ and smooth functions
There is, in general, a on-to-one correspondence between closed subalgebras of $C^*(X)$ (the algebra of bounded continuous real-valued functions) and the compactifications of $X$. Closed in the sense ...
4
votes
Accepted
Are compactifications of completely $T_{4}$ spaces completely $T_{4}$?
The obvious (to me) counterexamples are $\beta \mathbb{N}$ and $\beta \mathbb{R}$ ( the Čech-Stone compactifications) which are non-completely normal compactifications (classic fact, see ...
4
votes
Accepted
Does a flat compactification always exist?
This already fails if $S$ is regular of dimension $3$ and $\pi$ is quasi-finite. Indeed, let $X$ be a normal affine variety over $\mathbf C$ of dimension $3$ with an isolated non-Cohen–Macaulay ...
4
votes
Accepted
How complicated can the path component of a compact metric space be?
Not every path connected separable metric space is homeomorphic to a path component of a compact metric space. The following cardinality arguments can be used:
Fact 1. There is up to homeomorphism ...
4
votes
Accepted
Are degrees and ramification degrees preserved upon passing to the smooth compactification?
The first claim is true. The degree can be defined as the degree of the extension on function fields $k(C_1)/k(C_2)$, but $k(\tilde{C}_1) = k(C_1)$ and $k(\tilde{C}_2)= k(C_2)$ so $k(C_1)/k(C_2) = k(\...
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 ...
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