connect2 - 2 Hence what we have shown is if G = A B is a...

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2 Hence, what we have shown is: if G = A B is a separation of G , then each set U G lies entirely inside either A or B . But, by assumption, there is a non-empty common intersection for all sets U G – say x U for all U G . Then either x A or x B ; again, wlog, assume x A . Hence, U A = for any U G , which means (since each such U is contained in either A or B ) that U A for each U G . In other words, G A . But G = A B , and A B A B = ; this means that B = . This contradicts our choice of B . The point of Lemma 5 is that we can now speak of the biggest connected set containing the point x . That is: Definition 7. Let ( X, d ) be a metric space, and let x C . Define G x to be the collection of all connected subsets of X that contain the point x . In this case, x G x , and hence the intersection is non-empty. Therefore, by Lemma 5, G x is connected. The connected set C x G x is called the connected component of X containing x .
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