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

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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 definition 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 finitely many components. Then its components are clopen. Proof. Proposition 14 shows that the components are closed, so we need to show they are open. But as we noted in the remark following Lemma 9,
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