18
votes
Accepted
Is there a compact, connected, totally path-disconnected topological group?
(I'm assuming the groups to be Hausdorff to avoid the discussion degenerate into idle banter.)
The answer is yes: $\{1\}$ is such a group.
The answer to the intended question (which is probably ...
- 63.9k
15
votes
Lorenz attractor path-connected?
The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.
The situation is somewhat similar to the topologist's sine curve: the graph of
$$...
- 28k
10
votes
Accepted
Is every path connected $F_\sigma$ subset of a plane an image of $[0,1)$?
No, this fails even for compact subsets of $\mathbb R^2$. Namely, let $X=C\times[0,1]\cup[0,1]\times\{0\}$, where $C$ is the Cantor set. It is clearly path connected. $X$ cannot be an image of $[0,1)$,...
- 28.2k
6
votes
Connectedness of Quot schemes
The simplest example when connectedness fails is the Hilbert scheme of lines on a quadric surface:
$$
\mathrm{Hilb}_{1+t}(\mathbb{P}^1 \times \mathbb{P}^1) = \mathbb{P}^1 \sqcup \mathbb{P}^1,
$$
where ...
- 39.3k
5
votes
topological group that is connected and locally connected but not path-connected
A locally compact example (actually compact abelian) is the Pontryagin dual $G$ of the discrete abelian group $\mathbf{Z}^\omega$.
Indeed
a compact abelian group is connected iff its Pontryagin ...
- 63.9k
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 ...
- 770
4
votes
Accepted
Is every metric continuum almost path-connected?
Yes there are such examples.
Assume the sequence $\varepsilon_n$ is very fast converging to $0$.
Consider a sequence of short $\varepsilon_n$-crooked maps between intervals $\mathbb{J}_n\to \mathbb{J}...
- 45k
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 ...
- 1,947
4
votes
Accepted
Inscribing a "chain" into an open cover
The answer is ``yes'' if $X$ is Hausdorff.
We can identify the arc $C$ with the unit interval $[0,1]$ and assume that $[0,1]$ is contained in $X$ and is covered by a family $\mathcal U$ of open ...
- 41.8k
4
votes
Accepted
Does $(\omega, E)$ with the cycle condition have an $\omega$-path?
The answer to both questions in negative.
Let $G$ be the graph on $\omega$ with $nE0$ and $nE1$ for all $n\geq 2$ (and no other edges). Clearly all $m\neq n$ with $n,m\geq 2$ are on a cycle of length $...
- 3,011
3
votes
A topological characterization of trees?
The answer seems to be yes. Just take a path between $p_1$ and $p_2$, and then, on each next $i$th step ($i\geq 3$), add a path from $p_i$ to already constructed space (that is, you take a path from $...
- 23.7k
3
votes
Does $(\omega, E)$ with the cycle condition have an $\omega$-path?
No. Draw edges from $0$ and $1$ to all numbers $n>1$. Now any two nodes lies on a cycle of length $4$. But there is no injective $\omega$-walk, since every edge touches either $0$ or $1$, and ...
- 236k
3
votes
Accepted
Does there always exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and …?
This is a standard result that every rectifiable curve in a metric space admits an arc length parametrization. The proof can be found in many sources. For examples Theorem 3.2 in
Hajłasz - Sobolev ...
- 28k
3
votes
Accepted
Riemannian manifolds: every compact subset is contained in a connected relatively compact open subset
You will need to assume $K$ lies in a connected component of $M$.
Then for every point $x\in K$ there is a small open balls in $M$ whose closure is compact (you can just do this in charts). Since $K$ ...
- 7,583
3
votes
Accepted
Connectedness of the set having a fixed distance from a closed set 2
Here is a picture of the set $F$ (in red) in my comment above (the black lines represent the sphere of radius 2). There is a component of $A$ inside the sphere and a component outside the sphere.
- 23.2k
3
votes
Accepted
Is there a homeomorphism between the sets of Schur stable and Hurwitz stable matrices in companion forms?
Essentially you're asking if there is a homeomorphism between the monic real polynomials of degree $n$ with roots in the open unit disk and those with roots in the left half plane. Just take $$\prod_{...
- 54.2k
3
votes
Accepted
Can we classify the general linear maps such that for a fixed matrix $A \in M_n(\mathbb R)$ the conjuagation action fixes the first column?
One can not say that $\mathcal E$ is always connected.
For $n=3, $(and similarly $n>3$) let $A$ be a matrix with $a_{i1}=1,\quad \forall i \in \{1,2,\ldots,n\} $
Then $\mathcal E$ ...
- 356
2
votes
Find all paths on undirected Graph
Just list them, brute force.
For every edge $(A,X_i)$ define a subgraph by deleting that edge. In the subgraph, recursively enumerate all paths starting at $X_i$, then for every such path add the edge ...
- 809
2
votes
Accepted
Simple closed curves in a simply connected domain
Let $B_r(y)\subset\mathbb C$ denote the open disk of radius $r$ and center $y\in\mathbb C$, and $B_r:=B_r(0)$. Let $h:B_1\to U$ a homeomorphism (e.g. a Riemann mapping) . Then $\Gamma_r:=h(\partial ...
- 60.5k
2
votes
Planar compact connected set whose boundary has a finite length is arcwise connected
The answer is positive. First suppose that
$K=\partial K$. I will show that in this case
$K$ is the image of a curve.
By definition of Hausdorff measure, for every $\epsilon>0$ we can cover $K$ ...
- 91.8k
1
vote
Accepted
Product of adjacency matrices and connected graphs
(If we allow loops and multiple edges with the corresponding entry of the
adjacency matrix being the number of such loops or edges, then any
non-negative matrix may be interpreted as an adjacency ...
- 256
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(\...
- 3,599
1
vote
Accepted
A plane ray which limits onto itself
It seems like there are decomposable rays which limit onto themselves.
Let $E=\bigcup_{n\geq0}E_n$ and $O=\bigcup_{n\geq0}O_n$, with $E_n=[2n,2n+1]$ and $O_n=[2n+1,2n+2]$. We want to construct a ray $...
- 10.6k
1
vote
Connectedness of Quot schemes
Let $X$ be an irreducible variety, $E$ a locally free sheaf on $X$, and $n \geq 0$ an integer. The Quot scheme $\textrm{Quot}_X(E,n)$ is connected by Thm. 1.4 here. To achieve disconnectedness, ...
- 430
1
vote
Opposite-nearest neighbor algorithm vs. nearest neighbor algorithm
To answer your question, YES, if the length of an edge is the same in
each direction, then starting at a vertex $y_1$ and picking the next edge of maximum length
that visits a new vertex will return a ...
- 1,042
1
vote
Accepted
Gromov Hausdorff distance to tubular neighborhood
I think I have figured this out. More specifically, it should hold that
$$
d_{GH}(M, M_\sigma) \leq \max\left\{2\sigma, \left(\frac{\epsilon}{s-2\sigma}-1\right)(\mathrm{diam}(M)+2\sigma)+\epsilon\...
- 71
1
vote
Is the set $\{\zeta: \rho(A(\zeta))< 1\}$ connected for matrices under parameterization of first $m$ rows?
As a test case, for $n=2, m=1$ one can see it is connected.
Consider a matrix $A=\left[\begin{array}{cc}a & b \\c & d\end{array}\right]$, and assume that $c, d$ are given so $\zeta_0=[c,d]$. ...
- 68.8k
1
vote
Proof of existence and uniqueness of solution to f(c)=0
Put $\mathbb{R}^n_0=\{x\in\mathbb{R}^n:\sum_ix_0=0\}$, and let $\pi\colon\mathbb{R}^n\to\mathbb{R}^n_0$ be the orthogonal projection. You have a map $f\colon(0,\infty)^n\to\mathbb{R}^n$, and we can ...
- 56.9k
1
vote
Is the function $g$ always injective where $g$ is obtained by lipschitz re-parametrization
Yes, but not because of the reason in the accepted answer. If you remove the portions where $g$ is constant, the length stays the same, you're not making it shorter. Instead, if there was such an ...
- 111
1
vote
Accepted
Is the function $g$ always injective where $g$ is obtained by lipschitz re-parametrization
It is injective because if $g(s)=g(t)$, $s<t$, then you can remove the interval $[s,t]$ from the domain of definition of $g$ and make the curve shorter. Then you rescale the domain to be $[0,1]$. ...
- 28k
Only top scored, non community-wiki answers of a minimum length are eligible