380: Accumulation Value of Net with Directed Index Set Is Convergence of Subnet

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A description/proof of that accumulation value of net with directed index set is convergence of subnet

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of convergence of net with directed index set.
• The reader knows a definition of subnet of net with directed index set.
• The reader knows a definition of accumulation value of net with directed index set.

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Target Context

• The reader will have a description and a proof of the proposition that any accumulation value of any net with directed index set is the convergence of a subnet.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any directed set, $S_1$, any topological space, $T$, any net with directed index set, $f_1:S_1\to T$, and any accumulation value of $f_1$, $p\in T$, there is a subnet of $f_1$, $f_1\circ f_2:S_2\to T$ where $S_2$ is a directed set and $f_2:S_2\to S_1$, such that $p$ is a convergence of $f_1\circ f_2$.

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2: Proof

Let us define $S_2$ as $\{(\alpha ,N_p)|\alpha \in S_1,N_p\text{ is any neighborhood of }p\text{ such that }{f}_1(\alpha )\in N_p\}$ with the relation such that $({\alpha }_1,N_{p-1})\le ({\alpha }_2,N_{p-2})$ if and only if ${\alpha }_1\le {\alpha }_2$ and $N_{p-2}\subseteq N_{p-1}$, where there is at least 1 such $N_p$ for each $\alpha$, because there is a neighborhood of $p$ and there is a neighborhood of $f_1(\alpha )$ and the union of the 2 neighborhoods is a one ($N_p$ does not need to be connected). Note that all the possible pairs are included in $S_2$.

$S_2$ is indeed a directed set, because 1) $({\alpha }_1,N_{p-1})\le ({\alpha }_1,N_{p-1})$; 2) if $({\alpha }_1,N_{p-1})\le ({\alpha }_2,N_{p-2})$ and $({\alpha }_2,N_{p-2})\le ({\alpha }_3,N_{p-3})$, $({\alpha }_1,N_{p-1})\le ({\alpha }_3,N_{p-3})$; 3) for any $({\alpha }_1,N_{p-1})$ and $({\alpha }_2,N_{p-2})$, there is a $({\alpha }_3,N_{p-3})$ such that $({\alpha }_1,N_{p-1})\le ({\alpha }_3,N_{p-3})$ and $({\alpha }_2,N_{p-2})\le ({\alpha }_3,N_{p-3})$, because by the definition of directed set, there is an ${\alpha }_4$ such that ${\alpha }_1\le {\alpha }_4$ and ${\alpha }_2\le {\alpha }_4$ and as $p$ is an accumulation value, there is an ${\alpha }_3$ such that ${\alpha }_4\le {\alpha }_3$ and $f_1({\alpha }_3)\in N_{p-1}\cap N_{p-2}:=N_{p-3}$, then ${\alpha }_1\le {\alpha }_3$ and ${\alpha }_2\le {\alpha }_3$ and $N_{p-3}\subseteq N_{p-1}$ and $N_{p-3}\subseteq N_{p-2}$.

Let us define $f_2$ as $(\alpha ,N_p)\mapsto \alpha$, which is final, because for any $\alpha$, $\alpha \le f_2((\alpha ,N_p))=\alpha$.

$p$ is a convergence of $f_1\circ f_2$, because for any neighborhood of $p$, $N_{p-0}$, there is a $({\alpha }_0,N_{p-0})$ because $p$ is an accumulation value of $f_1$, and for every $({\alpha }_0,N_{p-0})\le (\alpha ,N_p)$, $f_1\circ f_2((\alpha ,N_p))=f_1(\alpha )\in N_{p-0}$, because $f_1(\alpha )\in N_p\subseteq N_{p-0}$.

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References

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