12
votes
Accepted
Is there a Bernstein subset $X$ of $\mathbb{R}$ such that no continuous map $f : X → [0,1]$ is surjective?
No, if $X$ is a Bernstein set then there is a continuous surjection $X \rightarrow [0,1]$.
To see this, note that there is a continuous surjection $f: \mathbb R \rightarrow [0,1]$ such that the ...
9
votes
Proving the inequality involving Hausdorff distance and Wasserstein infinity distance
EDIT: answer 2 below is completely false, as pointed out by the OP. However this is such a typical example of wishful thinking that I believe it is worth leaving for the posterity. (I'll record it ...
6
votes
Does anyone use non-sober topological spaces?
As suggested by @DavidWhite, I am turning my comment into an answer.
One class of very naturally appearing non-sober spaces is that of Alexandroff topologies on infinite posets. An irreducible closed ...
Community wiki
5
votes
Accepted
Is there a metric separable space with the following properties...?
Let $X$ be a Bernstein subset of $\mathbb{R}$, so $X$ and its complement intersect every uncountable closed set in $\mathbb{R}$.
Let $f:X\to\mathbb{R}$ be continuous and assume $f[X]$ is uncountable. ...
4
votes
Accepted
In a topological group, is $G/A\to G/B$ a covering map if $A$ is open in $B$?
Let $C$ be the intersection of the conjugates $bAb^{-1}$. Assume the additional hypothesis that $C$ is open in $B$. This holds, for example, when $B$ is locally connected.
The set $B\backslash C$, ...
2
votes
Accepted
Factorization systems for vector bundles
Normally, we consider vector bundles over some base space $X$. The simplest case is if $X$ is a point. Then the category of vector bundles over $X$ is just the category of vector spaces (over whatever ...
2
votes
Accepted
Quotients in categories of metric spaces
$\newcommand{cr}{\operatorname{cr}}$ Start by letting $X_T$ be any second-countable, functionally Hausdorff space which is not regular. Edit: We'll need to add an additional assumption here, the ...
2
votes
The cardinality of projections of subsets of the Hilbert cube by inner products
Consistently, there is no such set.
Given $X\subset [0,1]^{\mathbb N}$ of cardinality continuum it suffices to find $z\in(0,1)$ such that $x\mapsto \sum_n x_n z^n$ is injective on some subset $X'\...
1
vote
Let K be a compact set in a surface, U component of S-K, K'=S-U. K has finitely many components. Does every component of K' contains a component of K?
Since $S$ is locally connected, every connected component of the open set $S\setminus K$ is open in $S$. Denote the connected components other than $U$ as $\{V_i\}_{i\in I}$. Thus $S=K\sqcup U\sqcup(\...
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