7
votes
Accepted
Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$?
Such sets are called triods. R. L. Moore (Concerning triods in the plane and the junction points of plane continua, Proceedings of the National Academy of Sciences USA, vol. 14, 1928, pp. 85-88) ...
- 8,573
5
votes
Stone-Čech compactification
As for Q1, the answer is no. As remarked by Narutaka, the hyperstonean cover of $[0,1]$ does not have isolated points (Corollary 2.22) and all points of a discrete space $\Gamma$ are isolated in $\...
- 10.9k
5
votes
When is the Minkowski sum of weighted compact sets $w_1 B_1 + w_2 B_2 + \ldots$ (with $w \in L^1$) closed?
Another case.
Let $B_1, B_2, \dots$ be compact sets in $\mathbb R^n$, all bounded by $R > 0$. Let $w_n$ be nonnegative numbers with sum $1$. The product topological space $X = B_1 \times B_2 \...
- 36.3k
3
votes
Open covering with bounded diameters
The answer is given by the Lebesgue's Covering Theorem (numbered as 1.8.20 in Engleking's book "Theory of dimensions: finite and infinite"): If $\mathcal F$ is a finite closed cover of the $...
- 32.9k
3
votes
Accepted
Distinguishing topologically weak topologies of Banach spaces
The weak topologies of the spaces $\ell_1$ and $L_1$ are not homeomorphic because of the following
Theorem. Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has ...
- 32.9k
2
votes
Under what general conditions is the set $S := \left\{\int_{X}v(x)\pi(x)\,\mathrm{d}P(x) \mid \pi: X \to A\right\}$ closed?
I guess that if $v$ is $P$-integrable then the answer is positive, and actually the set is compact.
Indeed, what you are looking for in this case is the compactness of the Aumann integral of the ...
- 1,639
2
votes
Accepted
Can one explore a surface along ‘piecewise planar’ curves?
Consider the set
$S=\{\,(t,t^2,t^3)\in\mathbb R^3\mid\,t\in\mathbb R\}$.
Clearly $S$ is connected, and so is its complement $A=\mathbb R^3\setminus S$.
Note that each plane has at most 3 points of ...
- 37.4k
2
votes
What is definition of branched covering?
At the top of page ten of this paper the authors write "the standard two-sheeted branched covering of the sphere by the torus, branched over four points which become the four boundary circles of ...
- 18.6k
1
vote
A neighborhood $Y$ of a set $X$ such that the line segment connecting any point in $Y$ and its projection to $X$ is contained in $Y$
This is to confirm the stronger claim mentioned in (1).
Set $X_r=X\cap \overline{B_r(0)}$ and $X^r=X\setminus B_r(0)$ (both are closed, for any $t$). Define
$$
f(t)=\min\left\{1,\sup\left\{r\colon ...
- 18.7k
1
vote
Accepted
Is there at least one path in the common boundary of two open sets?
The answer to the second question is yes: there is an arc containing uncountable points of $B\cap\partial C$. It is enough to prove it in the case $n=2$.
Applying an affine transformation if necessary,...
- 3,232
1
vote
Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$?
The question been answered in How many tacks fit in the plane?.
In fact Moore proved a multidimensional version of the theorem: only countably many sets homeomorphic to the $n$-dimensional disc with ...
- 22.5k
1
vote
Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$?
One sketch is as follows: For each Y-set, we can put a small open disk in the vertex, such that the three prongs divide the disk in three sectors.
In each sector, we can find a rational coordinate, ...
- 14.4k
1
vote
Accepted
Consistency of the Hurewicz dichotomy property
The anwer is in: Tall, Franklin D.; Todorcevic, Stevo; Tokgöz, Seçil, The strength of Menger’s conjecture.
In the paper they prove (among other things) that the Hurewicz dichotomy extended to all ...
- 803
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