lecture note /topology / lecturer :Zahir Dobeas AL- nafie

Limit Points and

1

If (X,  ) is a then it is usual to refer to the elements of the X as points.

Definition. Let A be a subset of a topological space (X,  ). A point x  X is said to be a 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 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 , 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 zX\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.