connect4 - 4 hand, E = A B, and so A B B; in particular, A...

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4 hand, E = A B , and so A B B ; in particular, A B . But A B = , and this is only possible if A = , contra our assumptions. Whence, A 1 ± = , and indeed, neither A 1 nor B 1 are empty. Therefore, E = A 1 B 1 is a separation of E , which is therefore disconnected as claimed. ± Proof of Proposition 14. As C is a connected component, it is non-empty. Fix any point x C – so C = C x . By Lemma 15, C x is also connected, and hence is a connected set containing x . But by the de±nition of C x , this means the connected set C x is included in the union that is C x , and hence C x C x . But any set is contained in its closure, and hence C x = C x , meaning that C = C x is closed, as advertised. ± In many cases, components provide examples of so-called clopen sets – sets that are both open and closed. Corollary 16. Let X be a metric space with Fnitely many components. Then its components are clopen. Proof.
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This note was uploaded on 05/06/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.

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