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The proof of this theorem is by contradiction. Suppose for convex sets S and T there are elements a and b such that a and b both belong to S∩T, i.e., a belongs to S and T and b belongs to S and T and there is a point c on the straight line between a and b that does not belong to S∩T. This would mean that c does not belong to one of the sets S or T or both. For whichever set c does not belong to this is a contradiction of that set's convexity, contrary to assumption. Thus no such c and a and b can exist and hence S∩T is convex.

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Q: Intersection of two convex set is convex?

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It can be if the set consists of convex shapes, for example.

the union of two convex sets need not be a convex set.

You normally do not have an intersection of only one set. The intersection of a set with itself is the set itself - a statement that adds little value. The intersection of two sets is the set which contains elements that are in each of the two sets.

I will prove a more general theorem from which your answer follows immediately. Theorem: The intersection of any number (including 2) of convex polygons is convex.ProofLet C be the intersection of Ci which is a set of iconvex polygons. By definition of intersection, if two points A and B belong to C then they belong to every one of the Ci . But the convexity of each of the Ci tells us that line segment AB is contained in Ci . Therefore, the line segment AB is in C and because ABwas arbitrary we conclude that C is convex

Given two or more sets there is a set which is their union and a set which is there intersection. But, there is no such thing as a "union intersection set", as required for the answer to the question.

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The intersection is the set of solutions that satisfy two or more mathematical expressions.

the other one is intersection

Their intersection.

An intersection is the region of space that forms when two forms overlap (the intersection of two lines makes a point, the intersection of two planes makes a line, etc.). In set theory, it is the set formed when two or more sets overlap in terms of common elements. With respect to roads, it is the place where two roads cross each other.

The answer depends on how it is halved. If the plane is divided in two by a step graph (a zig-zag line) then it will not be a convex set.

The intersection of two distinct planes is a line. The set of common points in the line lies in both planes.

The intersection of two sets S and T is the set of all elements that belong to both S and T.

No. It can be infinite, finite or null. The set of odd integers is infinite, the set of even integers is infinite. Their intersection is void, or the null set.

The concept of closure: If A and B are sets the intersection of sets is a set. Then if the intersection of two sets is a set and that set could be empty but still a set. The same for union, a set A union set Null is a set by closure,and is the set A.

Intersection.

Intersection

A point or set of numbers that is a solution to two equations.

The intersection of two or more mathematical objects is the set of all points that are common to all of them. In set theory, that would be the elements in common. In geometry, it would be the set of all points in common. For example, the intersection of two different planes is a line; the intersection of a plane and a cone are the conic sections: circle, ellipse, parabola and hyperbola.

It is used in set theory to indicate intersection. The intersection of two sets, A and B, is the set of all elements that are in A as well as in B.

That is called the intersection of the sets.

Intersection Ad Muncher

You need two sets to have an intersection. If you have two sets, call them R and S, then their intersection is the set T that contains all elements of R that also belong to S OR all elements of S and also belong to R...it's the same thing.

A convex polygon is one with no reflex angles (angles that measure more than 180 degrees when viewed from inside the polygon). More generally a convex set is on where a straight line between any two points in the set lies completely within the set.

no