376: For Hausdorff Topological Space, Net with Directed Index Set Can Have Only 1 Convergence

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A description/proof of that for Hausdorff topological space, net with directed index set can have only 1 convergence

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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 Hausdorff topological space.
• The reader knows a definition of convergence 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 for any Hausdorff topological space, any net with directed index set can have only 1 convergence at most.

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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, $S$, any Hausdorff topological space, $T$, and any net with directed index set, $f:S\to T$, $f$ can have only 1 convergence at most.

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2: Proof

Suppose $f$ had 2 convergences, $p_1,p_2\in T$. For any neighborhood of $p_j$ where $j=1\text{ or }2$, $N_j$, there would be an index, $i_j\in S$, such that $f(i)\in N_j$ for every $i\in S,i_j\le i$. As $T$ is Hausdorff, let us take $N_1$ and $N_2$ as disjoint. By the definition of directed set, there would be an index, $i_3\in S$, such that $i_1\le i_3$ and $i_2\le i_3$. For every $i_3\le i$, $f(i)\in N_1$ and $f(i)\in N_2$, which is a contradiction, as $N_1$ and $N_2$ would be disjoint.

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References

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