Editing Convex set (section)

=== Minkowski addition ===
{{Main|Minkowski addition}}
[[File:Minkowski sum graph - vector version.svg|thumb|alt=Three squares are shown in the nonnegative quadrant of the Cartesian plane. The square {{math|''Q''<sub>1</sub> {{=}} [0, 1] × [0, 1]}} is green. The square {{math|''Q''<sub>2</sub> {{=}} [1, 2] × [1, 2]}} is brown, and it sits inside the turquoise square {{math|1=Q<sub>1</sub>+Q<sub>2</sub>{{=}}[1,3]×[1,3]}}.|[[Minkowski addition]] of sets. The <!-- [[Minkowski addition|Minkowski]]&nbsp; -->[[sumset|sum]] of the squares&nbsp;Q<sub>1</sub>=[0,1]<sup>2</sup> and&nbsp;Q<sub>2</sub>=[1,2]<sup>2</sup> is the square&nbsp;Q<sub>1</sub>+Q<sub>2</sub>=[1,3]<sup>2</sup>.]]

In a real vector-space, the ''[[Minkowski addition|Minkowski sum]]'' of two (non-empty) sets, {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}}, is defined to be the [[sumset|set]] {{math|''S''<sub>1</sub>&nbsp;+&nbsp;''S''<sub>2</sub>}} formed by the addition of vectors element-wise from the summand-sets
<math display=block>S_1+S_2=\{x_1+x_2: x_1\in S_1, x_2\in S_2\}.</math>
More generally, the ''Minkowski sum'' of a finite family of (non-empty) sets {{math|''S<sub>n</sub>''}} is <!-- defined to be --> the set <!-- of vectors --> formed by element-wise addition of vectors<!--  from the summand-sets -->
<math display=block> \sum_n S_n = \left \{ \sum_n x_n : x_n \in S_n \right \}.</math>

For Minkowski&nbsp;addition, the ''zero set''&nbsp;{{math|{0} }} containing only the [[null vector|zero&nbsp;vector]]&nbsp;{{math|0}} has [[identity element|special importance]]: For every non-empty subset&nbsp;S of a vector space
<math display=block>S+\{0\}=S;</math>
in algebraic terminology, {{math|{0} }} is the [[identity element]] of Minkowski addition (on the collection of non-empty sets).<ref>The [[empty set]] is important in Minkowski addition, because the empty set annihilates every other subset: For every subset {{mvar|S}} of a vector space, its sum with the empty set is empty: <math>S+\emptyset=\emptyset</math>.</ref>