S26 - 1st December 2004 Munkres 26 Ex. 26.1 (Morten...

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1st December 2004 Munkres § 26 Ex. 26.1 (Morten Poulsen). (a) . Let T and T 0 be two topologies on the set X . Suppose T 0 ⊃ T . If ( X, T 0 ) is compact then ( X, T ) is compact: Clear, since every open covering if ( X, T ) is an open covering in ( X, T 0 ). If ( X, T ) is compact then ( X, T ) is in general not compact: Consider [0 , 1] in the standard topology and the discrete topology. (b) . Lemma 1. If ( X, T ) and ( X, T 0 ) are compact Hausdorff spaces then either T and T 0 are equal or not comparable. Proof. If ( X, T ) compact and T 0 ⊃ T then the identity map ( X, T 0 ) ( X, T ) is a bijective continuous map, hence a homeomorphism, by theorem 26.6. This proves the result. ± Finally note that the set of topologies on the set X is partially ordered, c.f. ex. 11.2, under inclusion. From the lemma we conclude that the compact Hausdorff topologies on X are minimal elements in the set of all Hausdorff topologies on X . Ex. 26.2 (Morten Poulsen). (a) . The result follows from the following lemma. Lemma 2. If the set X is equipped with the finite complement topology then every subspace of X is compact. Proof. Suppose A X and let A be an open covering of A . Then any set A 0 ∈ A will covering all but a finite number of points. Now choose a finite number of sets from A covering A - A 0 . These sets and A 0 is a finite subcovering, hence A compact. ± (b) . Lets prove a more general result: Let X be an uncountable set. Let T c = { A X | X - A countable or equal X } . It is straightforward to check that T c is a topology on X . This topology is called the countable complement topology. Lemma 3. The compact subspaces of X are exactly the finite subspaces.
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This note was uploaded on 01/12/2011 for the course MATH 110 taught by Professor Brown during the Fall '08 term at Arizona Western College.

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S26 - 1st December 2004 Munkres 26 Ex. 26.1 (Morten...

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