Topological and metric properties of convex sets #
We prove the following facts:
convex.interior: interior of a convex set is convex;convex.closure: closure of a convex set is convex;set.finite.compact_convex_hull: convex hull of a finite set is compact;set.finite.is_closed_convex_hull: convex hull of a finite set is closed;convex_on_norm,convex_on_dist: norm and distance to a fixed point is convex on any convex set;convex_on_univ_norm,convex_on_univ_dist: norm and distance to a fixed point is convex on the whole space;convex_hull_ediam,convex_hull_diam: convex hull of a set has the same (e)metric diameter as the original set;bounded_convex_hull: convex hull of a set is bounded if and only if the original set is bounded.bounded_std_simplex,is_closed_std_simplex,compact_std_simplex: topological properties of the standard simplex;
Alias of the reverse direction of real.convex_iff_is_preconnected.
Standard simplex #
Every vector in std_simplex 𝕜 ι has max-norm at most 1.
std_simplex ℝ ι is bounded.
std_simplex ℝ ι is closed.
std_simplex ℝ ι is compact.
Topological vector space #
If s is a convex set, then a • interior s + b • closure s ⊆ interior s for all 0 < a,
0 ≤ b, a + b = 1. See also convex.combo_interior_self_subset_interior for a weaker version.
If s is a convex set, then a • interior s + b • s ⊆ interior s for all 0 < a, 0 ≤ b,
a + b = 1. See also convex.combo_interior_closure_subset_interior for a stronger version.
If s is a convex set, then a • closure s + b • interior s ⊆ interior s for all 0 ≤ a,
0 < b, a + b = 1. See also convex.combo_self_interior_subset_interior for a weaker version.
If s is a convex set, then a • s + b • interior s ⊆ interior s for all 0 ≤ a, 0 < b,
a + b = 1. See also convex.combo_closure_interior_subset_interior for a stronger version.
If x ∈ closure s and y ∈ interior s, then the segment (x, y] is included in interior s.
If x ∈ s and y ∈ interior s, then the segment (x, y] is included in interior s.
If x ∈ closure s and x + y ∈ interior s, then x + t y ∈ interior s for t ∈ (0, 1].
If x ∈ s and x + y ∈ interior s, then x + t y ∈ interior s for t ∈ (0, 1].
In a topological vector space, the interior of a convex set is convex.
In a topological vector space, the closure of a convex set is convex.
Convex hull of a finite set is compact.
Convex hull of a finite set is closed.
If we dilate the interior of a convex set about a point in its interior by a scale t > 1,
the result includes the closure of the original set.
TODO Generalise this from convex sets to sets that are balanced / star-shaped about x.
If we dilate a convex set about a point in its interior by a scale t > 1, the interior of
the result includes the closure of the original set.
TODO Generalise this from convex sets to sets that are balanced / star-shaped about x.
If we dilate a convex set about a point in its interior by a scale t > 1, the interior of
the result includes the closure of the original set.
TODO Generalise this from convex sets to sets that are balanced / star-shaped about x.
A nonempty convex set is path connected.
A nonempty convex set is connected.
A convex set is preconnected.
Every topological vector space over ℝ is path connected.
Not an instance, because it creates enormous TC subproblems (turn on pp.all).
Normed vector space #
The norm on a real normed space is convex on any convex set. See also seminorm.convex_on
and convex_on_univ_norm.
The norm on a real normed space is convex on the whole space. See also seminorm.convex_on
and convex_on_norm.
If s, t are disjoint convex sets, s is compact and t is closed then we can find open
disjoint convex sets containing them.
Given a point x in the convex hull of s and a point y, there exists a point
of s at distance at least dist x y from y.
Given a point x in the convex hull of s and a point y in the convex hull of t,
there exist points x' ∈ s and y' ∈ t at distance at least dist x y.
Emetric diameter of the convex hull of a set s equals the emetric diameter of `s.
Diameter of the convex hull of a set s equals the emetric diameter of `s.
Convex hull of s is bounded if and only if s is bounded.