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1 vote
Accepted

Does the topology of Wasserstein space $(\mathcal P_p (E), W_p)$ coincide with the initial topology induced by $\mathcal C_b(E) \cup \{g_p\}$?

A topology generated by countably many point-separating real functions is metrizable. To apply this here, it suffices to show that there is a countable family $\mathcal{G}$ of bounded real functions ...
1 vote

Inductive limit of $\mathbb R^n$s is Hausdorff and second countable?

A good reason why $\mathbb R^\infty_0$ is not metrizable (hence not first countable, a property that coincides with metrisability for Hausdorff TVS) is Baire’s theorem. You can follow this path: (1) A ...
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4 votes

Connected and locally connected, but not path-connected

All examples have to be fairly ugly, because while the category of topological spaces has examples of connected and locally connected spaces which are not path connected, many closely related ...
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2 votes
Accepted

Conditions that ensure the metric topology of $E$ coincides with the initial topology induced by a collection of real-valued functions on $E$

If $\mathcal{F}$ is countable, $\tau'$ is metrizable with a compatible metric $\rho$ given by $$\rho(x,y)=\sum_n 2^{-n}~|f_n(x)-f_n(y)|\wedge1$$ for some enumeration of $\mathcal{F}$. Since the ...
4 votes

"All retracts are closed" as separation axiom

Let $X$ be the rationals with their subspace topology, and $X^+=X\cup\{\infty\}$ be its one-point compactification. Because $X$ is not locally compact, $X^+$ is not $T_2$. The space $X^+$ has the ...
1 vote
Accepted

A question about a realcompact space and upper semicontinuous function

The following characterisation is well known. It can be found in Engelking's book as Theorem 3.11.10. Theorem: A Tychonoff space $X$ is realcompact if and only if for each $p\in\beta X\setminus X$ ...
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1 vote

Group actions and "transfinite dynamics"

Here is a partial answer extracted from a Twitter user's answer that works for actions with uniformly bounded finite orbits, which indeed solves a special case that initially motivated this question. ...
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3 votes

"All retracts are closed" and "all compacts are closed"

EDIT: This answer relied on an accepted answer elsewhere that has now been updated to remove an oversight. See my note below. First I need to prove that the Arens-Fort space $X$ is not compactly ...
11 votes

Is there an infinite topological space with only countably many continuous functions to itself?

The answer to this question is affirmative: there exists a countable Hausdorff space $X$ such that every continuous map $X\to X$ is either constant or the identity. Many such spaces are constructed in ...
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1 vote

Why are the selection principle $S_\text{fin}(\Lambda, \Omega)$ and $S_\text{fin}(\mathcal{O},\Lambda)$ impossible for nontrivial spaces?

This is because $S_1(A,B)$ implies $S_{fin}(A,B)$, which in turn implies the property "A choose B": Every element of A contains an element of B. Every reasonable space has an open cover that ...
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6 votes
Accepted

How many pairwise non-homeomorphic non-empty closed subsets of the Cantor set are there?

There are $2^{\aleph_0}$ different subsets of the Cantor set up to homeomorphism. There can't be more than $2^{\aleph_0}$ of them because any subset of the Cantor set is separable. To construct $2^{\...
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1 vote
Accepted

Classification of Polish spaces up to a $\sigma$-homeomorphism

There are indeed continuum many. See: Kihara, T., & Pauly, A. (2022). Point Degree Spectra of Represented Spaces. Forum of Mathematics, Sigma, 10, E31. doi:10.1017/fms.2022.7
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6 votes
Accepted

Is there an infinite topological space with only countably many continuous functions to itself?

A partial answer: the only place where Kannan and Rajagopalan use the inequality $(2^\kappa)^+<2^{2^\kappa}$ is in the application of the Theorem on page 121. That theorem is a consequence of ...
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2 votes
Accepted

Spaces satisfying a strong Cartan-Hadamard theorem

Note that Hilbert spaces (of all dimensions finite or infinite) are the only geodesic spaces with extendable geodesics which are flat in the sense of Alexandrov. Therefore $X$ has to have extendable ...
1 vote

Are there infinitely many "generalized triangle vertices"?

I found this property of X(1138) no later than 2015. I called them X(4),X(74) and X(1138) system centers and tried to find another one but failed. I also found they all lie on the Neuberg's cubic ...
3 votes
Accepted

Extending a partially defined metric on a metrizable space

Here is a counterexample to Q2, with your stated extra condition. Let $X$ consist of the half-open unit interval $(0,1]$ on the $x$-axis in the plane, together with the full unit interval $[0,1]$ at ...
4 votes
Accepted

A stronger version of paracompactness

Lemma: Let $(U_{\alpha})_{\alpha\in A}$ be a finitely intersecting open cover of a space $X$. Then there is some partition $P$ of $A$ where $(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$ is a partition of ...

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