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Unit 6: Closed Sets and Limit Point
Theorem 1: Let X be a topological space. Then the following conditions hold: Notes
(a) and X are closed.
(b) Arbitrary intersections of closed sets are closed.
(c) Finite unions of closed sets are closed.
Proof:
(a) and X are closed because they are the complements of the open sets X and , respectively.
A
(b) Given a collection of closed sets , we apply De Morgan’s law,
J
X A X A .
J J
Since the sets X A are open by definition, the right side of this equation represents an
arbitrary union of open sets, and is thus open. Therefore, A is closed.
(c) Similarly, if A is closed for i = 1, …, n, consider the equation
i
n n
X A X A .
i i
i 1 i 1
The set on the right side of this equation is a finite intersection of open sets and is therefore
open. Hence A is closed.
i
Theorem 2: Let Y be a subspace of X. Then a set A is closed in Y if and only if it equals the
intersection of a closed set of X with Y.
Proof: Assume that A = C Y, where C is closed in X. Then X C is open in X, so that
(X C) Y is open in Y, by definition of the subspace topology. But (X C) Y = Y A. Hence
Y A is open in Y, so that A is closed in Y. Conversely, assume that A is closed in Y. Then Y A
is open in Y, so that by definition it equals the intersection of an open set U of X with Y. The set
X U is closed in X and A = Y (X U), so that A equals the intersection of a closed set of X with
Y, as desired.
Example 3: Let (Y, U) (X, T) and A Y.
Then A is U-closed iff A = F Y for some T closed set F.
or
A is U-closed iff A is the intersection of Y and a T-closed F.
Solution: Let (Y, U) (X, T) and A Y, i.e. (Y, U) is subspace of (X, T).
To prove that A is U-closed iff
A = F Y for some T-closed set F.
A is U-closed Y A is U-open.
Then Y A can be expressed as:
Y A = G Y for some T-open set G.
From which A = Y G Y = X Y G Y
= (X G) Y
= F Y, where F = X G is a T-closed set.
This completes the proof.
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