360: For Topological Space, Intersection of Basis and Subspace Is Basis for Subspace

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A description/proof of that for topological space, intersection of basis and subspace is basis for subspace

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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 topological space.
• The reader knows a definition of basis of topological space.
• The reader knows a definition of subspace topology.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, the intersection of any basis and any subspace is a basis for the subspace.

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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 basis, $B=\{B_{\alpha }\}$, and any subset, $S\subseteq T$, with the subspace topology, the intersection of the basis and the subset, $\{B_{\alpha }\cap S\}$, is a basis for the subspace.

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2: Proof

For any open set, $U\subseteq S$, on $S$, $U=U^{\prime }\cap S$ where $U^{\prime }\subseteq T$ is open on $T$. For any point, $p\in U$, as $p\in U^{\prime }$, there is a $B_{\alpha }$ such that $p\in B_{\alpha }\subseteq U^{\prime }$ by the definition of basis. As $p\in S$, $p\in B_{\alpha }\cap S\subseteq U^{\prime }\cap S=U$. So, for any open set, $U\subseteq S$, and any point, $p\in U$, there is a $B_{\alpha }\cap S$ such that $p\in B_{\alpha }\cap S\subseteq U$, which means that $\{B_{\alpha }\cap S\}$ is a basis for $S$ by the definition of basis.

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References

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