Manuscript En08-A@1 · En08-A.tex · proposition · c688d76d

Proposition: Interval basis criterion

Alexandrov intervals and manifold structure › A sufficient basis criterion · lines 1744–1762 · proof included or linked

En08-A@1#prop:basis

Suppose:

every point belongs to an interval;
if $a_{1} ≪ x$ and $a_{2} ≪ x$ , there exists $a$ with $a_{1} ≪ a, a_{2} ≪ a, a ≪ x$ ;
if $x ≪ b_{1}$ and $x ≪ b_{2}$ , there exists $b$ with $x ≪ b, b ≪ b_{1}, b ≪ b_{2}$ .

Then the intervals form a basis.

Proof. The first condition gives coverage. If $x$ lies in the intersection of two intervals, the other conditions produce $a, b$ with $x \in I (a, b)$ contained in that intersection, by transitivity.