how to prove the closure of a set is closed??

Answer 1

Let A be a set (in a topological space), C its closure and C' the complement of C. If x is in C', then there's a neighborhood U of x that does not intersect A. Since U is a neighborhood of each of its elements, this same condition satisfied by x is equally satisfied by every y of U. By the definition of closure, it follows every y of U is in C' and we conclude U is a subset of C', so that x is an interior point of C'. Since this holds for every x of C', it follows C' is open and C is therefore closed.

Note: Here the definition of closure is as the set of all closure points. We say x is a closure point of A if every neighborhood of x intersects A. Many authors define C as the intersection of all closed sets that contains A (that is, the smallest closed set that contains A). The 2 definitions are equivalent. If we take the second one, then the conclusion is immediate, because the intersection of any collection of closed sets is closed.

Answer 2

Let S be a set and C be its closure. Pick a point x in the closure of C, and any neighborhood U of x. Then U contains points of C. Pick any point y in C intersect U. Pick any neighborhood V of y. Since y is in C, V contains points of S. Next, U intersect V contains y so it is nonempty. But U intersect V is a neighborhood of y, and y is in C, so U intersect V contains points of S. Since U intersect V is a subset of U, U contains points of S. Thus any neighborhood of x contains points of S, hence x is in C. Hence the closure of C is a subset of C, thus is equal to C.