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Math Help - Connectedness and continuity

  1. #1
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    Connectedness and continuity

    The following is a theorem from my book, but i don't think that i follow the proof given:

    Theorem:
    Every continuous image of a connected metric space is connected.

    Proof:
    Suppose X and Y are metric spaces and g : X -> Y is continuous.

    Suppose g(X) is not a connected space.

    Then there exists a continuous function f from g(X) onto {0,1}. (i don't think i follow this)

    This yields (fog)(X) = f(g(X)) = {0,1}, and since fog is continuous,

    X does not satisfy the continuity criterion for connectedness and therefore is not connected. (I don't understand how this happens ? why is it not satisfying the continuity criterion ?)

    Any help will be appreciated.
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  2. #2
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    Re: Connectedness and continuity

    Suppose g(X) is not connected. Then, g(X) = U \cup V where U\cap V = \emptyset, and both U and V are open and nonempty. Define f:g(X) \to \{0,1\} by f(x) = \begin{cases}0 & \text{if }x \in U \\ 1 & \text{if }x \in V\end{cases}. Now, f is continuous if the preimage of open sets are open. So, let's look at the open sets of \{0,1\}. They are \{\emptyset, \{0\}, \{1\}, \{0,1\}\}. The preimage of the empty set is the empty set, which is open in g(X). The preimage of \{0\} is U, which is open in g(X). The preimage of \{1\} is V, which is open in g(X), and the preimage of \{0,1\} is g(X), which is obviously open in g(X). So, f is continuous. You have already seen the proof that the composition of continuous functions is continuous. So, (f\circ g):X \to \{0,1\} must be a continuous function. Hence, (f\circ g)^{-1}(\{0\}), (f\circ g)^{-1}(\{1\}) must be disjoint, nonempty, open subsets of X whose union is X. This implies X is not connected.

    Edit: Disjoint comes from the definition of a function. If they were not disjoint, then let x \in (f\circ g)^{-1}(\{0\}) \cap (f\circ g)^{-1}(\{1\}). Then (f\circ g)(x) = 0 and (f\circ g)(x) = 1 implying (f\circ g) is not even a function.

    Nonempty comes from the fact that U,V are nonempty subsets of g(X), so g^{-1}(U) and g^{-1}(V) must be nonempty subsets of X.

    Open comes from continuity of (f\circ g).

    The union (f\circ g)^{-1}(\{0\}) \cup (f\circ g)^{-1}(\{1\}) = (f\circ g)^{-1}(\{0,1\}) = X because the preimage of the entire image of a function is the entire domain.
    Last edited by SlipEternal; November 5th 2013 at 08:29 PM.
    Thanks from mrmaaza123
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