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Unformatted text preview: 11 Proof. If S is bounded, then by the Bolzano-Weierstrass Theorem (Proposition 2.3.7 ) every sequence in S has a convergent subsequence, and if S is also closed, then the limit of this subsequence must also be a point of S . Conversely, if S is not bounded , it cannot be sequentially compact since there must exist a sequence s k of points in S with b s k b > k ; such a sequence has no convergent subsequence. Similarly, if S is not closed , there must exist a convergent sequence s k of points in S whose limit L lies outside S ; since every subsequence also converges to L , S cannot be sequentially compact. With these deFnitions, we can formulate and prove the following. 11 The property of being compact has a speciFc deFnition in very general settings; how-ever, in the context of R 3 , this is equivalent to either sequential compactness or being closed and bounded....
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
- Fall '08