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Zahir Dobeas AL- Nafie lecture note /topology / lecturer :Zahir Dobeas AL- nafie Limit Points and Closure 1 If (X, ) is a topological space then it is usual to refer to the elements of the set X as points. Definition. Let A be a subset of a topological space (X, ). A point x X is said to be a limit point (or accumulation point or cluster point) of A if every open set, U, containing x contains a point of A diferent from x. Example. Consider the topological space (X, ) where the set X = {a,b,c,d,e}, the topology = {X, Ø, {a}, {c,d}, {a,c,d}, {b,c,d,e}}, and A = {a,b,c}. Then b, d, and e are limit points of A but a and c are not limit points of A. Proof. The point a is a limit point of A if and only if every open set containing a contains another point of the set A. So to show that a is not a limit point of A, it sufces to find even one open set which contains a but contains no other point of A. The set {a} is open and contains no other point of A. So a is not a limit point of A. The set {c,d} is an open set containing c but no other point of A. So c is not a limit point of A. To show that b is a limit point of A, we have to show that every open set containing b contains a point of A other than b. We shall show this is the case by writing down all of the open sets containing b and verifying that each contains a point of A other than b. The only open sets containing b are X and {b,c,d,e} and both contain another element of A , namely c. So b is a limit point of A. The point d is a limit point of A, even though it is not in A. This is so since every open set containing d contains a point of A. Similarly e is a limit point of A even though it is not in A.£ 2 Let (X, ) be a discrete space and A a subset of X. Then A has no limit Example. points, since for each x X, {x} is an open set containing no point of A diferent from x. Example. Consider the subset A = [a,b) of R. Then it is easily verified that every element in [a,b) is a limit point of A. The point b is also a limit point of A. Example. Let (X, ) be an indiscrete space and A a subset of X with at least two elements. Then it is readily seen that every point of X is a limit point of A. (Why did we insist that A have at least two points?) The next proposition provides a useful way of testing whether a set is closed or not. Proposition. Let A be a subset of a topological space (X, ). Then A is closed in (X, ) if and only if A contains all of its limit points. Proof. We are required to prove that A is closed in (X, ) if and only if A contains all of its limit points; that is, we have to show that (i) if A is a closed set, then it contains all of its limit points, and (ii) if A contains all of its limit points, then it is a closed set. Assume that A is closed in (X, ). Suppose that p is a limit point of A which belongs to X \ A. Then X \ A is an open set containing the limit point p of A. Therefore X \ A contains an element of A. This is clearly false and so we have a contradiction to our supposition. Therefore every limit point of A must belong to A. Conversely, assume that A contains all of its limit points. For each z X \A, our assumption implies that there exists an open set U z 3 z such that U z A = Ø; that is, U z X \ A. Therefore X \A = S zX\A U .z( Check this!) So X \A is a union of open sets and hence is open. Consequently its complement, A, is closed. 3 Example. As applications of Proposition 3.1.6 we have the following: (i) the set [a,b) is not closed in R, since b is a limit point and b / [a,b); (ii) the set [a,b] is closed in R, since all the limit points of [a,b] (namely all the elements of [a,b]) are in [a,b]; (iii) (a,b) is not a closed subset of R, since it does not contain the limit point a; (iv) [a, ) is a closed subset of R. the set of all Proposition. Let A be a subset of a topological space (X, ) and A 0 the set of all limit points of A. Then A A 0 is a closed set. From Proposition 3.1.6 it sufces to show that the set A A 0 contains all of its limit Proof. points or equivalently that no element of X \ (A A 0) is a limit point of A A 0. Let p X \ (A A 0). As p / A 0, there exists an open set U containing p with U A = {p} or • . But p / A, so U A = Ø. We claim also that U A 0 = Ø. For if x U then as U is an open set and U A = Ø, x / A 0. Thus U A 0 = Ø. That is, U (A A ) = Ø, 0and p U. This implies p is not a limit point of A A 0 and so A A 0 is a closed set. Definition. Let A be a subset of a topological space (X, ). Then the set A A0 consisting of A and all its limit points is called the closure of A and is denoted by A. Remark. It is clear from Proposition 3.1.8 that A is a closed set. By Proposition 3.1.6 and Exercises 3.1 #5 (i), every closed set containing A must also contain the set A 0. So A A 0 = A is the smallest closed set containing A. This implies that A is the intersection of all closed sets containing A. .
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