= Midpoint test for a closed set
Checking only <midpoints> is not enough for an arbitrary <subset> of a real <vector space>. The <rational numbers> contain the <midpoint> of every pair of <rational numbers>, but the <line segment> from $0$ to $1$ also contains <irrational numbers>. For instance $\sqrt2/2$ lies on that <line segment> and is not rational. This shows why the definition quantifies over every real $t\in[0,1]$.
There is a useful additional hypothesis: a <closed subset> $C\subseteq\mathbb R^d$ containing the <midpoint> of every pair of its points is a convex set. Repeated bisection includes all points $(1-t)x+ty$ with <dyadic rational> $t\in[0,1]$. These coefficients approach every real $t\in[0,1]$; being a <closed subset> then includes each <limit>. This is a short proof outline, and the <rational numbers> example identifies precisely why a <closed subset> was required.
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