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Math Help - proving a map continuous between topological spaces

  1. #1
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    proving a map continuous between topological spaces

    I want to show that the function:

    f: Z x R -> R define by f(z,x) = x+z

    is continuous.

    In this case R is endowed with the euclidean topology and Z is the set of integers as a group with the + (normal addition) operator has the discrete topology.

    So an open set in U \subset R looks like some union of open balls. The preimage of any such set looks like countably infinitely many copies of these sets shifted by integer amounts.

    Now I know that the product topology is the coarsest topology which makes the projection functions continuous.

    Can we apply this fact to show the set f^{-1}(U) is in our product topology? How?


    Or is it easier than that? Can we show that the preimage of U under f is open because it is the union of open sets in the product topology?

    Is it clear that the set {0} x U is open in the product topology? If so then the preimage is the union over z \in Z of {z} x U.
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  2. #2
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    Quote Originally Posted by iknowone View Post
    I want to show that the function:

    f: Z x R -> R define by f(z,x) = x+z

    is continuous.

    In this case R is endowed with the euclidean topology and Z is the set of integers as a group with the + (normal addition) operator has the discrete topology.

    So an open set in U \subset R looks like some union of open balls. The preimage of any such set looks like countably infinitely many copies of these sets shifted by integer amounts.

    Now I know that the product topology is the coarsest topology which makes the projection functions continuous.

    Can we apply this fact to show the set f^{-1}(U) is in our product topology? How?


    Or is it easier than that? Can we show that the preimage of U under f is open because it is the union of open sets in the product topology?

    Is it clear that the set {0} x U is open in the product topology? If so then the preimage is the union over z \in Z of {z} x U.
    In your problem an open set of a product topology on \mathbb{Z} \times \mathbb{Re} has a form U \times V, where U is an open set of a discrete topology on \mathbb{Z} and V is an open set of a Euclidean topology on \mathbb{Re}.

    Let W be an open set in \mathbb{Re}, which is not equal to \mathbb{Re} itself . We shall show that f^{-1}(W) is open in the product topology on \mathbb{Z} \times \mathbb{Re}. Since W is an open set in the Euclidean topology on \mathbb{Re}, it is a union of open intervals. Thus, W=\bigcup_{i} (a_i, b_i), where a_i < b_i. Then, f^{-1}(W) has the form \{z\} \times V, where V=\bigcup_{i} (a_i-z, b_i-z), where a_i < b_i . Thus, f^{-1}(W) is open in the product topology on \mathbb{Z} \times \mathbb{Re}.

    For the remaining cases,
    f^{-1}(\mathbb{Re}) = \mathbb{Z} \times \mathbb{Re}, which is open in the product topology on \mathbb{Z} \times \mathbb{Re}.
    f^{-1}(\emptyset) = \emptyset, which is open in the product topology on \mathbb{Z} \times \mathbb{Re}.

    Thus, f is continuous.
    Last edited by aliceinwonderland; September 16th 2009 at 03:10 AM. Reason: Addition: line 3. "which is not equal to R itself".
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