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S17 - 1st December 2004 Munkres 17 Ex 17.3 A B is closed...

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1st December 2004 Munkres § 17 Ex. 17.3. A × B is closed because its complement ( X × Y ) - ( A × B ) = ( X - A ) × Y X × ( Y - B ) is open in the product topology. Ex. 17.6. (a) . If A B , then all limit points of A are also limit points of B , so [Thm 17.6] A B . (b) . Since A B A B and A B is closed [Thm 17.1], we have A B A B by (a). Conversely, since A A B A B , we have A A B by (a) again. Similarly, B A B . Therefore A B A B . This shows that closure commutes with finite unions. (c) . Since A α A α we have A α A α by (a) for all α and therefore A α A α . In general we do not have equality as the example A q = { q } , q Q , in R shows. Ex. 17.8. (a) . By [Ex 17.6.(a)], A B A and A B B , so A B A B . It is not true in general that A B = A B as the example A = [0 , 1), B = [1 , 2] in R shows. (However, if A is open and D is dense then A D = A ). (b) . Since A α A α we have A α A α for all α and therefore A α A α . (In fact, (a) is a special case of (b)). (c) . Let x A - B . For any neighborhood of x , U - B is also a neighborhood of x so U ( A - B ) = ( U - B ) A ( U - B ) A = since x is in the closure of A [Thm 17.5]. So x A - B . This shows that A - B A - B . Equality does not hold in general as R - { 0 } = R - { 0 } R - { 0 } = R . Just to recap we have (1) A B A B ( A B , B closed A B ) (2) A B = A B (3) A B A B ( A D = A if D is dense.) (4) A α A α (5) A α A α (6) A - B A - B Dually, (1) A B Int A Int B ( A B , A open A Int B ) (2) Int ( A B ) = Int A Int B (3) Int ( A B ) Int A Int B These formulas are really the same because X - A = X - Int A, Int ( X - A ) = X - A Ex. 17.9. [Thm 19.5] Since A × B is closed [Ex 17.3] and contains A × B , it also contains the closure of A × B [Ex 17.6.(a)], i.e. A × B A
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