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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) $d(x,y)+d(y,z) \geq d(x,z)$.
0
votes
Accepted
Points of differentiability of squared distance from a point in metric spaces
Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the $\ell_1$-metric $d:X\times X\to\mathbb R$ defined by
$$d\big((x,i),(y,j)\big)=|x-y|+|i-j|.
$$
It seems that this metric spac …
6
votes
Which points in the Samuel compactification of a metric space $X$ are limits of uniformly di...
A metric space $X$ is called isometrically homogeneous if for any points $x,y\in X$ there exists a bijective isometry $f:X\to X$ such that $f(x)=y$.
For isomemtrically homogeneous spaces this proble …
3
votes
Quotient of compact metrizable space in Hausdorff space
For the Cantor starcase function $f:C\to[0,1]$ from the standard ternary Cantor set $C$ onto the interval $[0,1]$ and for the standard Euclidean metric $d$ on $C$ the quotient pseudometric $d_\sim$ is …
5
votes
Accepted
Does there exist a countable metric space which is Lipschitz universal for all countable met...
The affirmative answer to this problem follows from
Lemma. For any countable dense subsets $X,Y$ in the half-line $\mathbb R_+=[0,+\infty)$ there exists a $C^2$-smooth function $f:\mathbb R_+\to\math …
4
votes
Extending homeomorphisms between compact metric subsets
In the "positive" direction, there exist some deep results (called Z-set unknotting theorems) on extensions of homeomorphisms between Z-sets of Menger manifolds, see Theorem 4.1.18 in the book "Inver …
3
votes
0
answers
161
views
A characterization of Cauchy filters on countable metric spaces?
Given a filter $\mathcal F$ on a countable set $X$, consider the family
$$\mathcal F^+:=\{A\subset X:\forall F\in\mathcal F\;(A\cap F\ne\emptyset)\}.$$
The following characterization is well-known.
…
5
votes
Accepted
Product topology from two premetric spaces induced by sum of premetrics?
The answer to this question is negative.
Consider the subspace $M_1=\{0\}\cup\{\frac 1n+\tfrac{i}{nm}:n,m\in\mathbb N\}$ of the complex plane and the space $M_2=M_1\cup\{\frac1n:n\in\mathbb N\}$ endow …
3
votes
Accepted
Takahashi minimization theorem for lower pseudo-continuous functions on complete metric spaces
The Takahashi Theorem holds also for lower pseudo-continuous functions.
To derive a contradiction, assume that $f:X\to [0,+\infty]$ a proper lower pseudo-continuous function such that for any point $ …