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Thread: Subnet convergence

  1. #1
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    Subnet convergence

    Let $\displaystyle X$ be a topological space.
    Show that if a net $\displaystyle x_{i\ i \in I}$ in $\displaystyle X$ has an accumulation point $\displaystyle p$, then there is a subnet of $\displaystyle x_{i\ i \in I}$ that converge to $\displaystyle p$.

    I assume this is the equivalent of subsequence convergence in a metric space.. i'm having hard time applying the net convergence definitions.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aharonidan View Post
    Let $\displaystyle X$ be a topological space.
    Show that if a net $\displaystyle x_{i\ i \in I}$ in $\displaystyle X$ has an accumulation point $\displaystyle p$, then there is a subnet of $\displaystyle x_{i\ i \in I}$ that converge to $\displaystyle p$.

    I assume this is the equivalent of subsequence convergence in a metric space.. i'm having hard time applying the net convergence definitions.
    And what precisely are you having difficulties with? This follows directly from the definition, so perhaps if you could narrow down which part of that definition troubles you we could help you better.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    And what precisely are you having difficulties with? This follows directly from the definition, so perhaps if you could narrow down which part of that definition troubles you we could help you better.
    since $\displaystyle p$ is an accumulation point. the set $\displaystyle I_U$={ $\displaystyle i \in I $: $\displaystyle x_i \in U$ , $\displaystyle U$ is open } is co-final in $\displaystyle I$. therefore I can take the set $\displaystyle J$={ $\displaystyle I_U$ : $\displaystyle p \in U$, $\displaystyle U$ is open } with the reverse inclusion.
    The net $\displaystyle x_{j\ j \in J} $ clearly converge to $\displaystyle p$.
    is this a subnet? if so how can show that? did i mess up the definitions?
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