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Thread: Topology

  1. #1
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    Topology

    Hi.

    A fast question (i dont want a proof):

    Let $\displaystyle (X_1, T_1)$ and $\displaystyle (X_2, T_2)$ two topological spaces. $\displaystyle (X_1, T_1)$ Itīs not Hausdorff.
    Let $\displaystyle fX_1, T_1)\to(X_2, T_2)$ a function continuous. Its possible that $\displaystyle (X_2, T_2)$ 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 $\displaystyle (X_1, T_1)$ and $\displaystyle (X_2, T_2)$ two topological spaces. $\displaystyle (X_1, T_1)$ Itīs not Hausdorff.
    Let $\displaystyle fX_1, T_1)\to(X_2, T_2)$ a function continuous. Its possible that $\displaystyle (X_2, T_2)$ 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 $\displaystyle (X_1, T_1)$ and $\displaystyle (X_2, T_2)$ two topological spaces. $\displaystyle (X_1, T_1)$ Itīs not Hausdorff.
    Let $\displaystyle fX_1, T_1)\to(X_2, T_2)$ a function continuous. Its possible that $\displaystyle (X_2, T_2)$ to be Hausdorff?

    I think no.

    Thanks
    Take an example that $\displaystyle T_1$ is an indiscrete topology given on $\displaystyle X_1$ and $\displaystyle (X_2, T_2)$ 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 $\displaystyle (X_2, T_2)$ is any non-Hausdorff topological space and $\displaystyle T_1$ is a discrete topology (which is Hausdorff) given on $\displaystyle X_1$. All maps having $\displaystyle (X_1, T_1)$ as a domain and $\displaystyle (X_2, T_2)$ 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?

    $\displaystyle (X_1, T_1)$ not Hausdorff, $\displaystyle pX_1, T_1) \to (X_2, T_2)$ continuous and surjective, implies $\displaystyle (X_2, T_2)$ 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?

    $\displaystyle (X_1, T_1)$ not Hausdorff, $\displaystyle pX_1, T_1) \to (X_2, T_2)$ continuous and surjective, implies $\displaystyle (X_2, T_2)$ not Hausdorff.

    Thanks.
    Consider this counterexample.

    Let $\displaystyle X_2$ be the real line $\displaystyle \mathbb{Re}$ with the standard topology; let $\displaystyle X_1$ be the product topological space $\displaystyle X_2 \times Y_1$, where $\displaystyle Y_1$ is the $\displaystyle \mathbb{Re}$ with the indiscrete topology. Then $\displaystyle X_1$ is not a Hausdorff space but $\displaystyle X_2$ is a Hausdorff space. You can construct a surjective continuous map $\displaystyle f:X_2 \times Y_1 \rightarrow X_2$ given $\displaystyle f(x_2, y_1) = x_2$, where $\displaystyle x_2 \in X_2$ and $\displaystyle y_1 \in Y_1$.
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  6. #6
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    Quote Originally Posted by aliceinwonderland View Post
    Not necessarily true. Consider $\displaystyle (X_2, T_2)$ is any non-Hausdorff topological space and $\displaystyle T_1$ is a discrete topology (which is Hausdorff) given on $\displaystyle X_1$. All maps having $\displaystyle (X_1, T_1)$ as a domain and $\displaystyle (X_2, T_2)$ 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 $\displaystyle \mathbb{Re}$ with the discrete topology and $\displaystyle \mathbb{Re}$ 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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