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Math Help - Topology of pointwise convergence

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    Topology of pointwise convergence



    Not really sure how to begin this one. Any help appreciated!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    Not really sure how to begin this one. Any help appreciated!
    What do you mean topologized by the family? Give \mathbb{X} the weak topology generated by the family of maps?
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    I have no idea what that means. I assume its the intersection of all the open balls of radius r given by the family of semi norms. There's a lot about this concept out there, especially with respect to locally convex spaces, but I still don't really understand it.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post
    I have no idea what that means. I assume its the intersection of all the open balls of radius r given by the family of semi norms. There's a lot about this concept out there, especially with respect to locally convex spaces, but I still don't really understand it.
    So you're saying give it the metric topology where an open ball is \displaystyle B\left(x_0;r\right)=\bigcap_{x\in[0,1]}\left\{y\in\mathbb{X}:\rho_x(y-x_0)<r\right\}??
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    I'm guessing, although I believe that topology separates points, and I think the idea of the proof is to demonstrate te consequences of a non metrizable topological space, or something.
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    Quote Originally Posted by Drexel28 View Post
    So you're saying give it the metric topology where an open ball is \displaystyle B\left(x_0;r\right)=\bigcap_{x\in[0,1]}\left\{y\in\mathbb{X}:\rho_x(y-x_0)<r\right\}??
    It's not a metric topology, basically because there are uncountably many of the defining seminorms \rho_x. A basic family of neighbourhoods of f would consist of sets of the form \{g\in\mathbb{X}:\rho_{x_j}(f-g) = |f(x_j)-g(x_j)|<\varepsilon\; (1\leqslant j\leqslant n)\} for each finite set of points x_1,...,x_n\in[0,1]. It is called the topology of pointwise convergence because a sequence (or more generally a directed net) \{f_i\} converges to the limit  f iff f_i(x)\to f(x) for each point x\in[0,1].

    For the last part of the question, I think that you have to use a cardinality argument. You want to find a sequence (f_n) in \mathbb{X} which converges to 0 (in the topology of pointwise convergence) so slowly that if you multiply its terms by any sequence of scalars that tends to infinity then the resulting sequence (\gamma_nf_n) no longer converges to 0.

    The set \Gamma of all sequences of integers that tend to infinity has cardinality 2^{\aleph_0}, which is the same as the cardinality of [0,1]. So there exists a bijective map \phi:\Gamma\to [0,1]. Define the function f_n\in\mathbb{X} by f_n(\phi(\gamma)) = 1/\gamma_n\ (\gamma\in\Gamma).

    Then f_n(x)\stackrel{n}{\to}0 for all x\in[0,1], but \gamma_nf_n(\phi(\gamma)) = 1\stackrel{n}{\not\to}0.

    Finally, I leave you to check that if (\gamma_nf_n) fails to converge to 0 for all sequences of integers tending to infinity, then the same will be true for all sequences of reals tending to infinity.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Opalg View Post
    It's not a metric topology, basically because there are uncountably many of the defining seminorms \rho_x. A basic family of neighbourhoods of f would consist of sets of the form \{g\in\mathbb{X}:\rho_{x_j}(f-g) = |f(x_j)-g(x_j)|<\varepsilon\; (1\leqslant j\leqslant n)\} for each finite set of points x_1,...,x_n\in[0,1]. It is called the topology of pointwise convergence because a sequence (or more generally a directed net) \{f_i\} converges to the limit  f iff f_i(x)\to f(x) for each point x\in[0,1].

    For the last part of the question, I think that you have to use a cardinality argument. You want to find a sequence (f_n) in \mathbb{X} which converges to 0 (in the topology of pointwise convergence) so slowly that if you multiply its terms by any sequence of scalars that tends to infinity then the resulting sequence (\gamma_nf_n) no longer converges to 0.

    The set \Gamma of all sequences of integers that tend to infinity has cardinality 2^{\aleph_0}, which is the same as the cardinality of [0,1]. So there exists a bijective map \phi:\Gamma\to [0,1]. Define the function f_n\in\mathbb{X} by f_n(\phi(\gamma)) = 1/\gamma_n\ (\gamma\in\Gamma).

    Then f_n(x)\stackrel{n}{\to}0 for all x\in[0,1], but \gamma_nf_n(\phi(\gamma)) = 1\stackrel{n}{\not\to}0.

    Finally, I leave you to check that if (\gamma_nf_n) fails to converge to 0 for all sequences of integers tending to infinity, then the same will be true for all sequences of reals tending to infinity.
    I understand, that was why I was asking this since the OP made it seems like he thought it was a metric topology. I'd like to mention that this topology is homeomorphic to [0,1]^{\mathbb{C}} with the product topology.
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