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Tagged with metric-spaces axiom-of-choice
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Is it possible to prove that any two points of a convex complete metric space are connected by some metric segment without the axiom of choice?
We say that a point $m$ is between points $p$ and $q$ of a metric space $(M, d)$ if $d(p, q) = d(p, m) + d(m, q)$ and $p ≠ m ≠ q$.
A metric space $M$ is said to be metrically convex if given any two ...
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