352: Criteria for Collection of Open Sets to Be Basis

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A description/proof of criteria for collection of open sets to be 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 1
• 2: Proof 1
• 3: Description 2
• 4: Proof 2

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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 admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of some criteria for any collection of open sets to be a 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 1

For any topological space, $T$, any collection of open sets, $\{B_{\alpha }\}$, is a basis of $T$, if and only if each open set on $T$ is the union of some elements of the collection.

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2: Proof 1

Suppose that each open set on $T$ is the union of some elements of the collection. For any open set, $U$, there is the union, ${\cup }_{{\alpha }_1}{B}_{{\alpha }_1}=U$, and any $B_{{\alpha }_1}$ satisfies $B_{{\alpha }_1}\subseteq U$, so, the collection is a basis.

Suppose that the collection is a basis. Any open set, $U$, is the union of some elements of the collection, because around any point, $p\in U$, there is an open set, $U_p\subseteq U$, by the local criterion for openness; there is an open set, $B_{\alpha }\subseteq U_p$, in the basis, by the definition of basis; $U$ is the union of such open sets in the basis.

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3: Description 2

For any set, $S$, and any collection of subsets, $B=\{B_{\alpha }\subseteq S\}$, the collection of all the unions of elements of $B$ (the empty set is included as the union of no element) constitutes a topology for $S$, and $B$ is a basis of the topological space, if and only if 1) $S$ is the union of all the elements of $B$, which is $S=\cup B_{\alpha }$, and 2) for each sets, $B_1,B_2\in B$, and each point, $p\in B_1\cap B_2$, there is a set, $B_3\in B$, such that $p\in B_3\subseteq B_1\cap B_2$.

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4: Proof 2

Suppose that the conditions 1) and 2) are satisfied. In the collection of all the unions of elements of $B$, $S$ is included because of 1) and the empty set is included, and any possibly uncountably infinite union of elements of the collection is included in the collection, because it is a union of elements of $B$, and any finite intersection of elements of the collection is included in the collection, because for any elements, ${\cup }_{\alpha }{B}_{\alpha }$ and ${\cup }_{\beta }{B}_{\beta }$, the intersection, $({\cup }_{\alpha }{B}_{\alpha })\cap ({\cup }_{\beta }{B}_{\beta })={\cup }_{\alpha ,\beta }(B_{\alpha }\cap B_{\beta })$, but because of 2), for any $p\in B_{\alpha }\cap B_{\beta }$, there is a $B_{\gamma }\in B$ such that $p\in B_{\gamma }\subseteq B_{\alpha }\cap B_{\beta }$, so, $B_{\alpha }\cap B_{\beta }=\cup B_{\gamma }$ as such $p$s cover $B_{\alpha }\cap B_{\beta }$, so, the intersection is $\cup B_{\gamma }$, which is in the collection. So, the collection is indeed a topology. $B$ is a basis of the topology, because for any open set, which is a union of elements of $B$, and any point in the open set, the point is in an element of $B$, and the element satisfies the definition of $B$'s being a basis.

Suppose that the supposed topology is really a topology and $B$ is a basis of the topological space. By Description 1, as $S$ is open, $S$ is the union of some elements of $B$, but then, $S$ is the union of all the elements of $B$, because adding any other remaining element, which is a subset of $S$, does not change the union. As $B$ is a basis, $B_1$ and $B_2$ are open, and $B_1\cap B_2$ is open, and by the definition of basis, there is such a $B_3$.

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References

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