Compact Sets and Hausdorff SpacesDate: 03/19/2003 at 19:29:18 From: Abby Subject: Compact Sets How do you prove that every compact subset of a metric space is closed? Date: 03/20/2003 at 09:10:40 From: Doctor Jacques Subject: Re: Compact Sets Hi Abby, Metric spaces have an important property which we shall use: given two distinct points a and b, we can find open sets U and V such that U contains a, V contains b, and U and V are disjoint. This is easy to see: if d(a,b) = s, consdider the open balls of centers a and b and radius s/3. Any topological space that verifies that property is called a Hausdorff space, and the theorem is true in all such spaces. Assume that K is a compact subset of a metric space S. This means that, from every open cover of K, we can extract a finite subcover of K. We want to show that K is closed, or, equivalently, that its complement is open. Let a be _any_ element of S \ K, the complement of K. For each point x in K, we can find open sets U_x and V_x such that: x is in U_x a is in V_x U_x and V_x are disjoint. As every x in K belongs to some U_x, the collection of U_x cover K, and we can extract from that set a finite collection of open sets: U_1, .... U_n whose union contains K. Let V_1, ... V_n be the corresponding "V" sets. Let C be the union of the U_i, and D be the intersection of the V_i. C and D are disjoint, because every element of C belongs to some U_i, and cannot therefore belong to the corresponding V_i. We also know that C contains K, and D contains a. D is the intersection of a _finite_ number of open sets, and is therefore open. Can you continue from here? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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