WebC(X) and the positive, totally finite, finitely additive set functions ,u defined on Ba (X). Using the Stone-Cech compactification gX of X and the Riesz-Markov Theorem, we, following Knowles, adopt the following notations. If q is a positive linear functional on C(X), let +(f) = +(f) be its unique extension to C(QX) where fE C(/QX) and f=f on X ...
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WebSep 5, 2024 · Exercise 3.9. E. 10. Prove the following about lines and line segments. (i) Show that any line segment in E n is a bounded set, but the entire line is not. (ii) Prove that the diameter of L ( a ¯, b ¯) and of ( a ¯, b ¯) equals ρ ( a ¯, b … Web(i) K- totally bounded and complete; (ii) K- compact. From the Hausdorff theorem it follows: Corollary3.1. Let K⊂ (X,ρ) and for any ε>0 there exists ε- net for K, which is totally bounded set. Then K- is also totally bounded. The following theorem provides a criteria of totally boundedness subsets from ℓp. p,k·kp), p≥ 1, following ...
WebCompactness and Totally Bounded Sets Theorem 5 (Thm. 8.16). Let A be a subset of a metric space (X,d). Then A is compact if and only if it is complete and totally bounded. … WebJan 18, 2011 · 3,134. 8. Of course, proving the other statement is easy. If Xi is compact for every i, then it is totally bounded, too. By the previous (yet unproved) result, a countable product X of totally bounded spaces is totally bounded in the metric D. Since X is complete under D (Theorem 43.4.), we conclude (by Theorem 45.1.) that X is compact.
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WebOct 23, 2009 · 1 Answer. Sorted by: 6. A metric (or uniform) space is compact if and only if is is totally bounded and complete. So a subset of a complete metric space is compact if and only if it is totally bounded and closed. Hence in a complete metric space, (bounded implies totally bounded) is equivalent to (bounded and closed implies compact), a property ...
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