S29 - April 21, 2006 Munkres 29 Ex. 29.1. Closed intervals...

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April 21, 2006 Munkres § 29 Ex. 29.1. Closed intervals [ a,b ] Q in Q are not compact for they are not even sequentially compact [Thm 28.2]. It follows that all compact subsets of Q have empty interior (are nowhere dense) so Q can not be locally compact. To see that compact subsets of Q are nowhere dense we may argue as follows: If C Q is compact and C has an interior point then there is a whole open interval ( a,b ) Q C and also [ a,b ] Q C for C is closed (as a compact subset of a Hausdorff space [Thm 26.3]). The closed subspace [ a,b ] Q of C is compact [Thm 26.2]. This contradicts that no closed intervals of Q are compact. Ex. 29.2. (a) . Assume that the product Q X α is locally compact. Projections are continuous and open [Ex 16.4], so X α is locally compact for all α [Ex 29.3]. Furthermore, there are subspaces U C such that U is nonempty and open and C is compact. Since π α ( U ) = X α for all but finitely many α , also π α ( C ) = X α for all but finitely many α . But C is compact so also π α ( C ) is compact. (b) . We have Q X α = X 1 × X 2 where X 1 is a finite product of locally compact spaces and X 2 is a product of compact spaces. It is clear that finite products of locally compact spaces are locally compact for finite products of open sets are open and all products of compact spaces are compact by Tychonoff. So X 1 is locally compact. X 2 is compact, hence locally compact. Thus the product of X 1 and X 2 is locally compact. Conclusion : Q X α is locally compact if and only if X α is locally compact for all α and compact for all but finitely many α . Example : R ω and Z ω + are not locally compact.
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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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S29 - April 21, 2006 Munkres 29 Ex. 29.1. Closed intervals...

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