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 ...
Will Brian's user avatar
  • 17.4k
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 ...
leo monsaingeon's user avatar
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 ...
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. ...
KP Hart's user avatar
  • 9,795
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 $B\backslash C$, ...
Tom Goodwillie's user avatar
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 ...
David White's user avatar
  • 29.4k
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 ...
Tyrone's user avatar
  • 5,016
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'\...
Colin McQuillan's user avatar
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(\...
KhashF's user avatar
  • 2,588

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