This is question 5 on page 83 in Chapter 13 of James R. Munkers’ Topology 2nd Edition textbook.

Basis Proof

What is being proved:

If $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by \(\mathcal{A}\) on \(X\), $\mathcal{T}_{\mathcal{A}}$, equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. Let \(\{\mathcal{T}_\alpha\}\) be an indexed family of all topologies that contain $\mathcal{A}$.

Proving \(\mathcal{T}_{\mathcal{A}} \subset \bigcap \{\mathcal{T}_\alpha\}\)

Let $A \in \mathcal{T}_{\mathcal{A}}$ as \(\mathcal{T}_{\mathcal{A}}\) is generated by it. If \(\mathcal{A}\) is a subcollection of every \(\mathcal{T}_\alpha\), then, by defintion (specifically proposition 2) of a topology, the union of the elements of \(\mathcal{A}\) must also be in every \(\mathcal{T}_\alpha\) as \(\mathcal{A}\) is a subcollection. Therefore, at minimum, \(\bigcap \{\mathcal{T}_\alpha\}\) must contain \(\mathcal{T}_{\mathcal{A}}\).

\[\mathcal{T}_\mathcal{A} \subset \bigcap \{\mathcal{T}_\alpha\}\]

Proving \(\bigcap \{\mathcal{T}_\alpha\} \subset \mathcal{T}_{\mathcal{A}}\)

Let \(U \in \bigcap \{\mathcal{T}_\alpha\}\) for any \(U\) in the intersection. Because \(\{\mathcal{T}_\alpha\}\) contains all topologies on \(X\) that contain \(\mathcal{A}\), \(\mathcal{T}_\mathcal{A} \in \{\mathcal{T}_\alpha\}\). Because the union of the elements of \(\mathcal{A}\) must be in every \(\mathcal{T}_\alpha\), then \(\bigcap \mathcal{T}_\alpha\) must contain only the collection containing the union of the elements of \(\mathcal{A}\). This is \(\mathcal{T}_\mathcal{A}\). Thus,

\[U \in \mathcal{T}_\mathcal{A}\]

Finally,

\[\bigcap \{\mathcal{T}_\alpha\} \subset \mathcal{T}_{\mathcal{A}}\]

Conclusion of Basis Proof

As these two collections are subcollections of each other,

\[\bigcap \{\mathcal{T}_\alpha\} = \mathcal{T}_{\mathcal{A}}\] \[\tag*{$\blacksquare$}\]

Subbasis Proof

What is being proved:

If \(\mathcal{S}\) is subbasis on \(X\), then the topology generated by \(\mathcal{S}\) on \(X\), \(\mathcal{T}_\mathcal{S}\), equals the intersection of all topologies on \(X\) that contain \(\mathcal{S}\). Let \(\{\mathcal{T}_\alpha\}\) be an indexed family of all topologies that contain $\mathcal{A}$.

There is also a basis \(\mathcal{B}\) generated from \(\mathcal{S}\) such that \(\mathcal{B}\) generates the same topology as \(\mathcal{S}\).

Prove \(\mathcal{T}_{\mathcal{S}} \subset \bigcap \{\mathcal{T}_\alpha\}\)

Let \(S \in \mathcal{T}_\mathcal{S}\) as \(\mathcal{T}_\mathcal{S}\) is generated by it. If \(\mathcal{B}\) is a subcollection of every \(\{\mathcal{T}_\alpha\}\), then, by definition of a topology, the union of the elements of \(\mathcal{B}\) must also be in every \(\mathcal{T}_\alpha\) as \(\mathcal{B}\) is a subcollection. Therefore, \(\bigcap \{\mathcal{T}_\alpha\}\) must contain \(\mathcal{T}_\mathcal{B}\).

\[\mathcal{T}_{\mathcal{S}} \subset \bigcap \{\mathcal{T}_\alpha\}\]

Prove \(\bigcap \{\mathcal{T}_\alpha\} \subset \mathcal{T}_{\mathcal{S}}\)

Let \(U \in \bigcap \{\mathcal{T}_\alpha\}\) for any \(U\) in the intersection. Because \(\{\mathcal{T}_\alpha\}\) contains all topologies on \(X\) that contain \(\mathcal{B}\), \(\mathcal{T}_\mathcal{S} \in \{\mathcal{T}_\alpha\}\). Because the union of the elements of \(\mathcal{B}\) must be in every \(\mathcal{T}_\alpha\), then \(\bigcap \{\mathcal{T}_\alpha\}\) must contain only the collection containing the union of the elements of \(\mathcal{B}\). This is \(\mathcal{T}_\mathcal{S}\), Thus \(U \in \mathcal{T}_\mathcal{S}\).

Subbasis Proof Conclusion

Because \(\mathcal{T}_{\mathcal{S}} \subset \bigcap \{\mathcal{T}_\alpha\}\) and \(\bigcap \{\mathcal{T}_\alpha\} \subset \mathcal{T}_{\mathcal{S}}\),

\[\mathcal{T}_{\mathcal{S}} = \bigcap \{\mathcal{T}_\alpha\}\] \[\tag*{$\blacksquare$}\]

Extras

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