well-defined sets are sets that can identify easily while not well-defined are those that cannot determined easily :)
Why is it important to be able to identify sets and set theory as related to business operations?
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.
a polygon is convex
When it is divisible by itself and one.
the union of two convex sets need not be a convex set.
well-defined sets are sets that can identify easily while not well-defined are those that cannot determined easily :)
Steven R. Lay has written: 'Convex sets and their applications' -- subject(s): Convex sets
look at it
Why is it important to be able to identify sets and set theory as related to business operations?
These terms describe polygons. To identify a polygon as convex, we draw a segment from any vertex to any other vertex. This segment cannot go outside of the polygon. Non-convex is concave. If we draw a segment from a vertex to any other vertex, at least one of the segments will go outside of the polygon.
Yes, in optimization problems, the feasible region must be a convex set to ensure that the objective function has a unique optimal solution. This is because convex sets have certain properties that guarantee the existence of a single optimum within the feasible region.
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.
a polygon is convex
You can bring them to a coin dealer but you will get more selling them on EBay. Most sets are not very valuable. you can easily Google the year and get the value.
So that police can identify them easily.
To easily identify which dog is which.