10
votes
Accepted
Are any embeddings $[0,1]\to\mathbb{R}^3$ topologically equivalent?
As igorf pointed out in the comment, the answer to the first question is 'no'. A quick counterexample is by looking into the complement of Fox-Artin arc, which is not simply connected. See figure ...
- 2,083
5
votes
Curves in the plane and their number of holes
Your assumption, if I understood it correctly is that the curve has only double self-crossings.
Consider your curve on the sphere (plane with infinite point added). It defines a cell decomposition of ...
- 91.8k
4
votes
"Thickening" an arc on a 2-manifold
Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set.
Proof 1: Find an arc $\alpha'$ so that $\gamma = \alpha \cup \alpha'$ is a Jordan curve (+). The Jordan/Schoenflies [...
- 28.1k
4
votes
Why are there no wild arcs in the plane?
The desired result appears in M.H.A. Newman's classic book, Elements of the Topology of Plane Sets of Points (2nd ed., 1951), as Theorem 14.5 in Chapter VI, on pg. 164:
Theorem 14.5: Every simple arc ...
- 311
4
votes
Jordan plane curve such that $\frac{d(g(x),g(y))}{d(x,y)}\to0$?
Sure. It is a standard fact that the von Koch snowflake $C$ can be parametrized by
$$ f\colon S^1 \rightarrow C$$
where
$$ C^{-1}|x-y|^{1/p} \leq |f(x)-f(y)| \leq C|x-y|^{1/p},$$
$p\in (1,2)$ is the ...
- 41
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
1
vote
Homeomorphism and boundary of a complementary component
The answer to Question 2 is negative: Let $Y$ be a continuum consisting of the segment $[-1,1]\times\{0\}$ to which a sequence of half circles $C_n$ with radius $\frac1n$ lying in the upper halfplane ...
- 770
1
vote
Homeomorphism and boundary of a complementary component
For the modified question, here is a counter-example:
First, note that the Cantor set $K$ is (topologically) homogeneous: the group of homeomorphisms acts transitively. (One way to see this is by ...
- 12.2k
1
vote
Accepted
Curves in the plane and their number of holes
Here's a slightly more hands-on point of view, using just the Jordan curve theorem (although of course in the end it does come down to the same thing as Euler's formula somehow, as described by Alex.)
...
- 6,548
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