195: Set of Neighborhood Bases at All Points Determines Topology

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A description/proof of that set of neighborhood bases at all points determines topology

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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 neighborhood basis at point.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any set of neighborhood bases at all points determines the topology.

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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 topological space, $T$, any set of neighborhood bases at all points, $\{B_p\}$ where $p\in T$, determines the topology, which means that there cannot be any different topology for the points set of $T$, for which $\{B_p\}$ is a set of neighborhood bases at all points.

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2: Proof

Let $T_1$ denote the topology of $T$ and suppose that there is another topology, $T_2$, for the points set of $T$, for which $\{B_p\}$ is a set of neighborhood bases at all points. Let us think of any open neighborhood, $U_p$, of any point $p\in T$, for $T_1$. For any $p^{\prime }\in U_p$, $U_p$ is a neighborhood of $p^{\prime }$ for $T_1$, so, there is a neighborhood, $U_{p^{\prime }}$, for $T_1$ from $B_{p^{\prime }}$ such that $U_{p^{\prime }}\subseteq U_p$, by the definition of neighborhood basis. But $U_{p^{\prime }}$ is also a neighborhood for $T_2$ by the supposition. So, with respect to $T_2$, as there is a neighborhood at any point of $U_p$ contained in $U_p$, $U_p$ is open, which means that any open set for $T_1$ is open also for $T_2$. By symmetry, any open set for $T_2$ is open also for $T_1$. So, $T_1=T_2$.

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References

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