228: Set of Subsets Around Each Point with Conditions Generates Unique Topology with Each Set Being Neighborhood Basis

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A description/proof of that set of subsets around each point with conditions generates unique topology with each set being neighborhood basis

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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
• 3: Note

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Starting Context

• The reader knows a definition of neighborhood basis at point.
• The reader admits the proposition that for any topological space, any set of neighborhood bases at all points determines the topology.

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Target Context

• The reader will have a description and a proof of the proposition that for any set, any set of some subsets around each point with certain conditions generates the unique topology with each set being a neighborhood basis.

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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 set, $S$, and any set, $S_p$, of some subsets around each point, $p\in S$, which means that each element of $S_p$ contains $p$, that satisfies the conditions: [ 1) $S_p$ is not empty; 2) for any $S_1,S_2\in S_p$, there is an $S_3\in S_p$ such that $S_3\subseteq S_1\cap S_2$; 3) for any $S_1\in S_p$, there is a subset around $p$, $p\in S_2\subseteq S_1$, such that for each $p^{\prime }\in S_2$, there is an $S_3\in S_{p^{\prime }}$ such that $S_3\subseteq S_2$], $\{S_p\}$ generates the unique topology on $S$ such that $S_p$ is a neighborhood basis at $p$.

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2: Proof

Let us define a topology such that any subset $S_1\subseteq S$ is open if and only if at each point, $p\in S_1$, there is an $S_2\in S_p$ such that $S_2\subseteq S_1$. It is really a topology, because $S$ is open as $S_p$ is not empty and any element of $S_p$ is contained in $S$; the empty set is vacuously open; any union of open sets, ${\cup }_{\alpha }{U}_{\alpha }$, is open, because at each point, $p\in {\cup }_{\alpha }{U}_{\alpha }$, $p\in U_{\alpha }$ for an $\alpha$, there is an $S_1\in S_p$ such that $S_1\subseteq U_{\alpha }$ as $U_{\alpha }$ is open, and $S_1\subseteq {\cup }_{\alpha }{U}_{\alpha }$; any finite intersection of open sets, ${\cap }_i{U}_i$, is open, because at each point, $p\in {\cap }_i{U}_i$, $p\in U_i$ for each $i$, there is an $S_i\in S_p$ such that $S_i\subseteq U_i$ for each $i$, ${\cap }_i{S}_i\subseteq {\cap }_i{U}_i$, but there is an $S_{1^{\prime }}\in S_p$ such that $S_{1^{\prime }}\subseteq {\cap }_i{S}_i$ by the condition 2), and $S_{1^{\prime }}\subseteq {\cap }_i{S}_i\subseteq {\cap }_i{U}_i$.

$S_p$ is a neighborhood basis at $p$, because each element, $S_1\in S_p$, is a neighborhood, because by the condition 3), $S_1$ contains an $p\in S_2$ that is open; for each neighborhood, $N_p$, of $p$, there is an open neighborhood, $U_p\subseteq N_p$, of $p$, and there is an $S_1\in S_p$ such that $S_1\subseteq U_p\subseteq N_p$, by the definition of the topology.

The topology is unique by the proposition that for any topological space, any set of neighborhood bases at all points determines the topology.

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3: Note

The condition 3) is necessary in order to guarantee that $S_p$ is a neighborhood basis at $p$, although the topology (which may not be the only possible topology) can be defined without it.

As an example that $S_p$ cannot be any neighborhood basis for any topology without the condition 3), let us think of $S=\mathbb{R}$, $S_p=\{(p-1,p+1)\}$. $\{S_p\}$ actually satisfies the conditions 1) and 2) where 2) is only $(p-1,p+1)\subseteq (p-1,p+1)\cap (p-1,p+1)=(p-1,p+1)$ as there is only 1 element in $S_p$. Then, in order for $S_p$ to be a neighborhood basis at $p$, $(p-1,p+1)$ has to be a neighborhood of $p$, which has to contain an open neighborhood of $p$, which has to contain an element of $S_p$, which can be only $(p-1,p+1)$, so, $(p-1,p+1)$ has to be the open neighborhood. Then, $(p-1,p+1)\cap (p+1-1,p+1+1)=(p,p+1)$ has to be an open neighborhood of $p+1/2$, which has to contain an element of $S_{p+1/2}$, which can be only $(p+1/2-1,p+1/2+1)=(p-1/2,p+3/2)$, which is not really contained in $(p,p+1)$. So, whatever topology we choose, $S_p$ cannot be any neighborhood basis.

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References

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