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Components

# Components - CONNECTED COMPONENTS Recall the definition of...

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CONNECTED COMPONENTS Recall the definition of connectedness (2.45 in [1]). Definition 1. Let ( X, d ) be a metric space. A subset E X is called disconnected (or sepa- rated ) if there exists nonempty A, B E such that E = A B , and A B = A B = . A subset is called connected if it is not disconnected. Example 2. In R with the Euclidean metric, consider the set E = [0 , 1] (2 , 3) . Since (2 , 3) = [2 , 3] , the strong condition [0 , 1] (2 , 3) = holds, which implies that the two subsets A = [0 , 1] and B = (2 , 3) form a separation of E . Hence E is not connected. Example 3. Again in R , consider the set E = [0 , 1) (1 , 2] . This set is also disconnected: setting A = [0 , 1) and B = (1 , 2] we have A = [0 , 1] which does not intersect (1 , 2] = B , while (1 , 2] = [1 , 2] does not intersect [0 , 1) = A . Thus, E is disconnected. In this case, A B is not empty (it contains the point 1 ). We might say that A and B are adjacent . But they are not connected. A person living in A cannot communicate with a person living in B without sending a message outside the universe E = A B . Example 4. Once more in R , take E = [0 , 2] . We can express E as a union of two non- intersecting pieces, for example E = [0 , 1) [1 , 2] . (Indeed, any set with at least two points can be divided into two non-empty pieces.) But in this case, [0 , 1) [1 , 2] is non-empty (it consists of the single point 1 ), and hence this does not constitute a separation of E . The point of Definition 1 is that, to be disconnected, a set must consist of (at least) two pieces that do not touch . In fact, the set [0 , 2] is connected, as proved in Theorem 2.47 in [1]: the connected subsets of R are precisely the intervals. The concept of connectedness can be used to naturally divide any metric space into a collection of pieces, called its components . To see how to do this, we need the following lemma. Lemma 5. Let G be any collection of connected subsets of a metric space ( X, d ) . Suppose that T G is not empty. Then S G is connected. Remark 6 . The assumptions of Lemma 5 are stronger than needed to deduce the conclu- sion. For example, all we really need to assume is that, for any two sets U, V G either U V or U V is non-empty (i.e. any two sets in G touch). It will then following that the union of all sets in G is connected. The proof of this stronger theorem (i.e. with a weaker hypothesis) is only slightly more complicated than the proof below; but we only need the statement of Lemma 5 in what follows, so we’ll stick with what we need. Proof. To produce a contradiction, let us suppose that S G is disconnected. That is, let A, B be two non-empty sets so that S G = A B , and A B = A B = . Let U be any set in G . Note that U A A and U B B (you should work out why this is true). Hence, since A B = , it follows that U A ( U B ) A B = , and similarly ( U A ) U B = . But U G so U S G = A B ; this means that ( U A ) ( U B ) = U .

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