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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) $d(x,y)+d(y,z) \geq d(x,z)$.

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Is the topology generated by this weaker notion of a metric necessarily metrisable?

For a loose metric $d$ as above, we can consider the function $$d_1(x,y):=\sup\{|d(x,z)-d(y,z)|;z\in X\}.$$ It is easy to verify that $d_1$ is a metric, and $d(x,y)\leq d_1(x,y)\leq\rho(d(x,y))$ for a …
Saúl RM's user avatar
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1 vote

Changing a metric to that 2 points have different distance

Yes, such a distance $d'$ exists. We can suppose $d(x,X)=d(y,X)=k$ for some $k>0$. We can define a new distance $d'$ by $d'(a,b)=d(a,b)+|d(x,a)-d(x,b)|$. This easily implies $d'(x,X)=2k$, however for …
Saúl RM's user avatar
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1 vote
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Lipschitz maps with Hölder inverse preserve the doubling property

If I have understood the definitions correctly, $f(K)$ need not be doubling. For example consider a map $f$ from $[0,1]$ to the Hilbert space $\mathbb{R}\times l^2$ defined in the following way. Let $ …
Saúl RM's user avatar
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10 votes
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Relationship between doubling constant of a metric space and of a metric measure space

Apart from the obvious counterexample of the measure being $0$, if $(X,d,m)$ is doubling in the sense of metric measure spaces it will be doubling in the sense of metric spaces. Consider a ball $B(x,r …
Saúl RM's user avatar
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