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Juan Atacama
Juan Atacama
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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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$. We say that a metric space has midpoints if there is at least one midpoint between every two of its points.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it (hint).

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it (hint).

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$. We say that a metric space has midpoints if there is at least one midpoint between every two of its points.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it (hint).

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Juan Atacama
Juan Atacama
  • 307
  • 1
  • 10

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it (proofhint of it).

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it.

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it (hint).

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Juan Atacama
Juan Atacama
  • 307
  • 1
  • 10

Is the statement that every convex complete metric space has midpoints equivalent to the axiom of dependent choice (DC)?

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 $p ≠ m ≠ q$. Furthermore, if the equality $d(p, m) = d(m, q)$ holds for $m$, we say that $m$ is a midpoint between $p$ and $q$.

We say that a metric space has midpoints if there is at least one midpoint between every two of its points. A metric space $M$ is said to be metrically convex if given any two points $p, q ∈ M$ there exists at least one point $m ∈ M$ such that $m$ is between $p$ and $q$.

It can easily be shown that the axiom $\mathsf{DC}_{\omega_1}$ implies the statement in my question. Does this statement also follow from the $\mathsf{DC}$ axiom, which is weaker? I don't need to prove the converse implication because I already have proof of it.