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It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$). In other words, the values $X=|x-a|+|y-b|$, $Y=|x-b|+|y-a|$ and $Z=|x-y|+|a-b|$ satisfy ulrtatriangle inequality. This inequality can be also thought as a discrete analog of CAT[−∞] inequality.

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$).

In other words, the values $$X=|x-a|+|y-b|,$$ $$Y=|x-b|+|y-a|,$$ $$Z=|x-y|+|a-b|$$ satisfy ulrtatriangle inequality. This inequality can be also thought as a discrete analog of CAT[−∞] inequality.

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le |x-a| + |x-b| + |y-a| + |y-b|$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$). This inequality can be also thought as a discrete analog of CAT[−∞] inequality.

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$). In other words, the values $X=|x-a|+|y-b|$, $Y=|x-b|+|y-a|$ and $Z=|x-y|+|a-b|$ satisfy ulrtatriangle inequality. This inequality can be also thought as a discrete analog of CAT[−∞] inequality.

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$ | x - y | + | a - b | \le |x-a| + |x-b| + |y-a| + |y-b|$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$). This inequality can be also thought as a discrete analog of CAT[−∞] inequality.