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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 …
Włodzimierz Holsztyński's user avatar
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) …
Włodzimierz Holsztyński's user avatar
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 $\ …
Włodzimierz Holsztyński's user avatar
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 $\ N$.
Włodzimierz Holsztyński's user avatar
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 …
Włodzimierz Holsztyński's user avatar