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A metric space is a pair $(X,d)$, where $X$ is a set and $d:X \times X \to \mathbb{R}$ satisfies the following conditions for all $x,y,z \in X$. (Symmetry) $d(x,y)=d(y,x)$. (Identity of Indiscernibles) $d(x,y)=0$ if and only if $x=y$. (Triangle Inequality) $d(x,y)+d(y,z) \geq d(x,z)$.

3 votes
Accepted

Does there always exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the s...

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 spa …
LSpice's user avatar
  • 12.9k
8 votes

Uniform density of Lipschitz maps is space of continuous function — for general metric spaces

While in general one cannot expect density of Lipschitz mappings (as pointed out by Nik Weaver), the following classical result is in the positive direction. You can find a proof in Heinonen's book Le …
LSpice's user avatar
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3 votes
Accepted

Monotonicity of doubling dimension

This is Lemma 9.6(i) in J. C. Robinson, Dimensions, embeddings, and attractors. Cambridge Tracts in Mathematics, 186. Cambridge University Press, Cambridge, 2011. In the proof the author says "it is …
Piotr Hajlasz's user avatar
5 votes
Accepted

Domains in $\mathbb{R}^n$ for which Hajlasz-Sobolev spaces and Sobolev Spaces are the same

Your argument is not correct. If a property $P$ fails for $Y$ and $X\subset Y$, it does not follow that it fails for $X$. For example $X=\{0\}\subset\mathbb{R}=Y$ but there are many properties true fo …
Piotr Hajlasz's user avatar
5 votes
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Reference: Hajlasz-Sobolev Spaces with Values in a Metric Space

The question is not stated in a very clear manner, but nevertheless, the answer is: no. Separability. The space $M^{1,p}(X,d,\mu)$ is not separable even if $X$ is the standard ternary Cantor set, $d$ …
Piotr Hajlasz's user avatar
17 votes

What is the structure preserved by strong equivalence of metrics?

Metrics are strongly equivalent if the identity mapping $Id:(X,d_1)\to (X,d_2)$ is bi-Lipschitz. They preserve the class of Lipschitz mappings. Roughly speaking classical topology deals with notions …
Piotr Hajlasz's user avatar
4 votes
Accepted

Continuous inclusion of metric spaces of smaller capacity

The answer is no. You can have $X$ path connected and $Y$ totally disconnected satisfying your condition. Then any continuous map from $X$ to $Y$ is necessarily constant so there are no injective maps …
Piotr Hajlasz's user avatar
1 vote
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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 $[0,1]$. Resc …
Piotr Hajlasz's user avatar