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Unformatted text preview: 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...
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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.
- Spring '09