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Results tagged with metric-spaces
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user 23141
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) $d(x,y)+d(y,z) \geq d(x,z)$.
14
votes
Uniform density of Lipschitz maps is space of continuous function — for general metric spaces
Let $X$ be the unit circle and let $Y$ be the Koch snowflake, both with euclidean metric inherited from $\mathbb{R}^2$. There is a continuous homeomorphism from $X$ onto $Y$, but there is no nonconsta …
- 42.8k
4
votes
Accepted
Effect of snowflaking on doubling constants
Suppose $(X,d)$ has doubling constant $\lambda$. This means that for every $r$, every $d$-ball of radius $2r$ can be covered by $\lambda$ many $d$-balls of radius $r$. With respect to the metric $d^\a …
- 42.8k
3
votes
Accepted
Lipschitz-free space of countable uniformly discrete metric space
This works if $X$ has finite diameter, but not in general. An easy way to see why not is to look at the elementary molecules $m_{xy}$, since $\|m_{xy}\|_{\rm AE} = d(x,y)$. The norm of the correspondi …
- 42.8k
4
votes
Existence of a Hölder-free space
As AIM_BLB says in the comments, every Holder space of exponent $\alpha < 1$ is a Lipschitz space with respect to the metric $d^\alpha$. So the answer is an immediate "yes". May I add that I discuss H …
- 42.8k
1
vote
Accepted
If a function follows another one's range order, can we say it follows some continuity prope...
No, let $X = \{0\} \cup \{\frac{1}{n}: n \in \mathbb{N}\}$, define $f(t) = -t$ for all $t \in X$, and define $g(t) = \begin{cases}0&t > 0\cr 1&t = 0\end{cases}$.
- 42.8k
8
votes
Accepted
Intersection of nested open ball in complete metric spaces is nonempty?
I think this statement is true.
Suppose we had a counterexample $\{B(x_i,r_i)\}_{i=1}^\infty$ satisfying condition (1) for some $\epsilon > 0$ but whose intersection was empty. Observe that any subse …
- 42.8k
3
votes
Accepted
Ultraproduct of metric spaces
No, it's not an isometry in general. Let the index set be $\mathbb{N}$ and let each $X_n = \mathbb{N}$ with its usual metric. Let $\mathcal{U}$ be any free ultrafilter on $\mathbb{N}$. For each $n$ de …
- 42.8k
23
votes
When does a metric space have "infinite metric dimension"? (Definition of metric dimension)
Well, if $M$ has a metric basis $\{b_1, \ldots, b_{n+1}\}$ then the map $$x \mapsto (d(x,b_1), \ldots, d(x,b_{n+1}))$$ is a continuous injection from $M$ into $\mathbb{R}^{n+1}$. So, for example, no i …
- 42.8k
6
votes
Accepted
Finitely isometrically persistent metric spaces
There is no such space with more than one point. Let $X$ be any metric space and suppose there exist $x,y,z \in X$ such that $$a = d(x,y) > {\rm max}(d(x,z), x(y,z)) = b.$$
That is, there is a triangl …
- 42.8k
11
votes
Accepted
Trasportation metric (AKA Earth-Mover's, Wasserstein, etc.) as "natural" / "induced"?
Okay, so $X$ is a finite metric space and $D(X)$ is the positive part of the unit sphere of $l^1(X)$. We can consider $X$ as sitting inside $D(X)$ by identifying a point $x \in X$ with the function th …
- 42.8k