Page 23 - DMTH503_TOPOLOGY

Unit 1: Topological Spaces

For this G, we also claim Notes
G  D(A) = 
Let y  G be arbitrary.

Now G is an open set containing ‘y’ s.t.
G  A = , showing that y  D(A).
 any y  G  y  D(A)

This shows G  D(A) = 
 G  A = , G  D(A) = 
Now, G  [A  D(A)] = G  A)  [G  D(A)]
=    = 

 G  X – A  D(A)
 any x  X – A  D(A)
 G  T with x  G s.t. G  X – A  D(A)

This proves that x is an interior point of X – A  D(A).
Since x is arbitrary point of X – A  D(A).
Hence, every point of X – A  D(A) is an interior point of X – A  D(A).

 X – A  D(A) is open.
i.e., A  D(A) is closed.
Theorem 11: Let (X, T) be a topological space and A  X, then A is closed iff A  A or A  D(A).
A subset A of X in a topological space (X, T) is closed iff A contains each of its limit points.
C
Proof: Let A be closed  A is open.
Let x  A C

C
then A is open set containing x but containing no point of A other than x. This shows that x is not
a limit point of A.
C
Thus, no point of A is a limit point of A. Consequently, every limit point of A is in A and
therefore A  A.
Conversely, Let A  A.

To show: A is closed.
C
Let x be an arbitrary point of A .
C
Then x  A  x  A  x  A and x  A ( A  A)
 x  A and x is not a limit point of A.
  an open set G such that x  G and G  A =   G  A C
 x  G  A C

C
C
 A is the nhd of each of its points and therefore A is open.
Hence A is closed.

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