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Math Help - Topology

  1. #1
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    Topology

    Hi.

    A fast question (i dont want a proof):

    Let (X_1, T_1) and (X_2, T_2) two topological spaces. (X_1, T_1) Itīs not Hausdorff.
    Let X_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that (X_2, T_2) to be Hausdorff?

    I think no.

    Thanks
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  2. #2
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    Quote Originally Posted by asub1 View Post
    Hi.

    A fast question (i dont want a proof):

    Let (X_1, T_1) and (X_2, T_2) two topological spaces. (X_1, T_1) Itīs not Hausdorff.
    Let X_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that (X_2, T_2) to be Hausdorff?

    I think no.

    Thanks
    Not true. Take the line with the double origin, and map it to the real line as the identity and send the second origin also to 0. The map is continuous and the reals are Hausdorff.

    It is true however if you swap the spaces around i.e. f cts and X_2 not Hausdorff implies that X_1 is not Hausdorff.
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  3. #3
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    Quote Originally Posted by asub1 View Post
    Hi.

    A fast question (i dont want a proof):

    Let (X_1, T_1) and (X_2, T_2) two topological spaces. (X_1, T_1) Itīs not Hausdorff.
    Let X_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that (X_2, T_2) to be Hausdorff?

    I think no.

    Thanks
    Take an example that T_1 is an indiscrete topology given on X_1 and (X_2, T_2) is a Hausdorff space such that f is a constant map. We see that f is continuous.

    Quote Originally Posted by Focus View Post
    It is true however if you swap the spaces around i.e. f cts and X_2 not Hausdorff implies that X_1 is not Hausdorff.
    Not necessarily true. Consider (X_2, T_2) is any non-Hausdorff topological space and T_1 is a discrete topology (which is Hausdorff) given on X_1. All maps having (X_1, T_1) as a domain and (X_2, T_2) as a codomain are continuos (You can easily check this by the definition of a continuous function).
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  4. #4
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    Hi!

    I understand your example aliceinwonderland, but ...

    what about this?

    (X_1, T_1) not Hausdorff, X_1, T_1) \to (X_2, T_2)" alt="pX_1, T_1) \to (X_2, T_2)" /> continuous and surjective, implies (X_2, T_2) not Hausdorff.

    Thanks.
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  5. #5
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    Quote Originally Posted by asub1 View Post
    Hi!

    I understand your example aliceinwonderland, but ...

    what about this?

    (X_1, T_1) not Hausdorff, X_1, T_1) \to (X_2, T_2)" alt="pX_1, T_1) \to (X_2, T_2)" /> continuous and surjective, implies (X_2, T_2) not Hausdorff.

    Thanks.
    Consider this counterexample.

    Let X_2 be the real line \mathbb{Re} with the standard topology; let X_1 be the product topological space X_2 \times Y_1, where Y_1 is the \mathbb{Re} with the indiscrete topology. Then X_1 is not a Hausdorff space but X_2 is a Hausdorff space. You can construct a surjective continuous map f:X_2 \times Y_1 \rightarrow X_2 given f(x_2, y_1) = x_2, where x_2 \in X_2 and y_1 \in Y_1.
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  6. #6
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    Quote Originally Posted by aliceinwonderland View Post
    Not necessarily true. Consider (X_2, T_2) is any non-Hausdorff topological space and T_1 is a discrete topology (which is Hausdorff) given on X_1. All maps having (X_1, T_1) as a domain and (X_2, T_2) as a codomain are continuos (You can easily check this by the definition of a continuous function).
    Yes sorry, you need the function to be surjective.
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  7. #7
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    Quote Originally Posted by Focus View Post
    Yes sorry, you need the function to be surjective.
    The counterexample I showed you still works even if the continuous function in the example is surjective.
    Consider an identity map between \mathbb{Re} with the discrete topology and \mathbb{Re} with the indiscrete or cofinite topology. Then the map is continuous and surjective, but the domain of the map is Hausdorff and the codomain of the map is non-Hausdorff.
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