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Math Help - Definition of Connectedness

  1. #1
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    Definition of Connectedness

    In my book, "Real Mathematical Analysis" by Charles Pugh, a subset S of a metric space M is disconnected if there exists a proper, clopen subset of Because if the proper, clopen subset of S is A, then S = A union A(compliment)

    But for example on this wikipedia page: Connectedness - Wikipedia, the free encyclopedia

    It says that S is not connected (so disconnected) if it is not the union of two disjoint non-empty open sets, which is basically saying that S has no proper, open set.


    Which is the correct definition?
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  2. #2
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    Quote Originally Posted by JG89 View Post
    In my book, "Real Mathematical Analysis" by Charles Pugh, a subset S of a metric space M is disconnected if there exists a proper, clopen subset of Because if the proper, clopen subset of S is A, then S = A union A(compliment)
    But for example on this wikipedia page: Connectedness - Wikipedia, the free encyclopedia
    It says that S is not connected (so disconnected) if it is not the union of two disjoint non-empty open sets, which is basically saying that S has no proper, open set.
    To confuse you even more here is a third definition given by R L Moore in c1920.
    Two nonempty sets are separated if neither one contains a point nor a limit point of the other.
    (Of course, Prof Moore did not say nonempty- he did not believe empty point sets existed.)
    A set is connected if and only if it is not the union of two separated sets.

    But here is the kicker: All three are equivalent. You ought to prove it.
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    Thanks! I'll get on that.
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  4. #4
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    Be careful, they are not exactly equivalent. The definition by Moore, quoted by Plato, is for connected sets. The other two definitions are for connected spaces. For example, A= [0, 1]\cup [2, 3] is not a connected set in R (with the "usual" topology), but the only "clopen" sets in R are the empty set and R itself.

    Of course, if you think of A, above, as a vector space in its own right, with the topology inherited from R, then [0,1] and [2,3] are "clopen" in A.
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    Halls, if I am studying metric spaces, should I use the clopen and open set definitions?
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    Quote Originally Posted by JG89 View Post
    I am studying metric spaces, should I use the clopen and open set definitions?
    It is save to say that in many textbooks where metric spaces are the central focus, a space is said to be connected if it is not the union of two nonempty disjoint open sets. If a metric space is not connected, then the two separating sets are both open and closed (clopen).
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  7. #7
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    Quote Originally Posted by Plato View Post
    You ought to prove it.
    Definition 1: A metric space  M is connected if there is not a proper, clopen subset of M.

    Definition 2: A metric space  M is connected if  M is not the union, of two disjoint, non-empty open sets.

    Proof that the two definitions are equivalent:

    First suppose that  M is not the union of two disjoint, non-empty open sets. We prove that it does not contain a proper, clopen subset. Assume that indeed it does,  A \subset M is clopen. Then  M = A \cup A^c and  A \cap A^c = \varnothing . Both  A and  A^c are clopen, and thus open, sets that are disjoint and non-empty (we know they are non-empty because they are assumed to be proper subsets), which contradicts the fact that M is not the union of two disjoint, non-empty open sets.

    Now we must prove the other direction. We will prove it in contrapositive form. Suppose that  M is the union of two disjoint, non-empty open sets,  A and  B . We will prove that one of them is clopen. Take either  A or  B , say  A . Suppose it is not closed. There exists a convergent sequence  a_n \rightarrow a of its elements whose limit is not in  A , and thus must be in  B . Remember that  B is open and so  \exists \delta > 0 : d(a, x) < \delta \Rightarrow x \in B . Note that there exists at least one point in  \{x \in M : d(a,x) < \delta \} that's from the sequence  a_n since  a_n \rightarrow a . But this contradicts the fact that  A \cap B  = \varnothing , and so  A is closed, and since it is also open, it is clopen.  A is also a proper subset, and so there exists a proper, clopen subset of  M . QED


    Is this fine?
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  8. #8
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    That proof works.
    In the first part you need to be sure that set A is a proper subset.

    As for the second part recall that the complement of an open set is closed.
    Therefore M\setminus A=B so the set B is both open and closed.
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