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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) $d(x,y)+d(y,z) \geq d(x,z)$.
8
votes
Accepted
Quotient of metric spaces
The following counter-example leaves no doubts :-) (this will answer also some other - more basic - potential similar questions too). This example is related to the Cantor set:
Let $\ X:=[0;1]\ $ wit …
20
votes
If all balls at $x$ and $y$ are isometric is there an isometry sending $x$ to $y$?
There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:
$$d(x\ y) = d(a\ b) = 1$$
$$d(x\ a) = d(y\ b) = 2$$
$$d(x\ b) = d(y\ a) = 3$$
$$d(x\ c) = d(y\ c) …
4
votes
Accepted
Is there any standard procedure to properly define a composite metric?
There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:
Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ …
1
vote
Metrization of spaces of functions
Perhaps the constant functions should form a topological subspace of $\ C(M , N)$ which is canonically isomorphic with $\ N$. This would eliminate the non-metrizable spaces $\ N$.
2
votes
Axiom of Choice and continuous functions
Look Ma, no axiom of choice!
THEOREM 0 Let $X$ be a compact space. Let $\Phi$ be a non-empty family of closed subsets of $X$, $F := \bigcap \Phi$, and $G\supseteq F$ an open subse …