Download Limit Points and Closure

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.
topology
Consider the topological space (X,

)
where the set X = {a,b,c,d,e}, the
= {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.
(X,
)
Let A be a subset of a topological space (X,
 ). Then A is closed in
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 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.
Proposition.
Let A be a subset of a topological space (X,
limit points of A. Then A  A 0 is a closed set.
 ) and A 0
the set of all
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.