1190: For Topological Space and Point, Intersection of Finite Number of Neighborhoods of Point Is Neighborhood of Point

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description/proof of that for topological space and point, intersection of finite number of neighborhoods of point is neighborhood of point

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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: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of neighborhood of point on topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$t$: $\in T$
$\{N_{t,1},...,N_{t,n}\}$: $N_{t,n}\in \{\text{ the neighborhoods of }t\text{ on }T\}$
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Statements:
${\cap }_{j\in \{1,...,n\}}{N}_{t,j}\in \{\text{ the neighborhoods of }t\text{ on }T\}$
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2: Proof

Whole Strategy: Step 1: let any open neighborhood of $t$ contained in $N_{t,j}$ be $U_{t,j}$; Step 2: see that ${\cap }_{j\in \{1,...,n\}}{U}_{t,j}$ is an open neighborhood of $t$ and $t\in {\cap }_{j\in \{1,...,n\}}{U}_{t,j}\subseteq {\cap }_{j\in \{1,...,n\}}{N}_{t,j}$.

Step 1:

For each $j\in \{1,...,n\}$, there is an open neighborhood of $t$ contained in $N_{t,j}$, $U_{t,j}$, by the definition of a definition of neighborhood of point on topological space.

Step 2:

${\cap }_{j\in \{1,...,n\}}{U}_{t,j}$ is an open neighborhood of $t$, as the intersection of the finite number of open subsets.

${\cap }_{j\in \{1,...,n\}}{U}_{t,j}\subseteq {\cap }_{j\in \{1,...,n\}}{N}_{t,j}$, because $U_{t,j}\subseteq N_{t,j}$.

So, $t\in {\cap }_{j\in \{1,...,n\}}{U}_{t,j}\subseteq {\cap }_{j\in \{1,...,n\}}{N}_{t,j}$ where ${\cap }_{j\in \{1,...,n\}}{U}_{t,j}$ is open.

So, ${\cap }_{j\in \{1,...,n\}}{N}_{t,j}$ is a neighborhood of $t$.

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References

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