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Math Help - Criteria for connectedness

  1. #1
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    Criteria for connectedness

    I have to prove that if, X is not the union of two disjoint non-empty closed subsets of itself, then,


    Either X is empty or the only continuous functions from X to the discrete space {0,1} are the two constant functions.


    Attempt at the proof:


    Assume that the first one is true, and let f:X->{0,1} be continuous.


    Then f inverse of zero and f inverse of one are both disjoint and closed in X.


    How do i proceed further ?
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  2. #2
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    Re: Criteria for connectedness

    I don't understand what you mean by "Assume that the first one is true". You should assume that X cannot be represented as the union of two disjoint non-empty closed subsets of itself. Then, also assume X is nonempty (so, either X is empty, in which case the proposition is true, or X is nonempty, and you are in this case) and prove that the only continuous functions from X to the discrete space \{0,1\} are the two constant functions. As you said, define f:X \to \{0,1\} as a continuous function. You learned that a function is continuous if the preimage of every closed set is closed. Since \{0\}^c = \{1\} and \{1\}^c = \{0\}, both singleton sets are clopen (both open and closed) subsets of \{0,1\}. Since f is continuous, f^{-1}(\{0\}) and f^{-1}(\{1\}) must both be closed. Suppose x \in f^{-1}(\{0\}) \cap f^{-1}(\{1\}). Then f(x) = 0 and f(x) = 1, implying f is not even a function. So, the preimages must be disjoint. For any x \in X, either f(x) = 0 or f(x) = 1, so the union of the preimages must be the entire set. Suppose f^{-1}(\{0\}) \neq \emptyset. Then by the initial assumption ( X cannot be represented by the union of two disjoint, nonempty, closed subsets of itself), it must be that f^{-1}(\{1\}) = \emptyset. The other case: f^{-1}(\{0\}) = \emptyset. By the second assumption, X is nonempty, so it must be that f^{-1}(\{1\}) \neq \emptyset. Hence, either f(X) = \{0\} or f(X) = \{1\}.
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