Homeomorphic Image of Sub-Basis is Sub-Basis

Theorem
Let $T_\alpha = \struct{X_\alpha, \tau_\alpha}$ and $T_\beta = \struct{X_\beta, \tau_\beta}$ be topological spaces.

Let $\SS_\alpha \subseteq \tau_\alpha$ be a sub-basis for $\tau_\alpha$.

Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.

Let $\SS_\beta = \set{\phi \sqbrk S : S \in \SS_\alpha}$.

Then:
 * $\SS_\beta$ is a sub-basis for $\tau_\beta$

Proof
By definition of homeomorphism:
 * $\forall U \subseteq X_\alpha : U \in \tau_\alpha \iff \phi \sqbrk U \in \tau_\beta$

By definition of sub-basis:
 * $\SS_\alpha \subseteq \tau_\alpha$

Hence:
 * $\SS_\beta \subseteq \tau_\beta$

Let $\ds \BB_\alpha = \set {\bigcap \FF: \FF \subseteq \SS_\alpha, \FF \text{ is finite} }$.

Let $\ds \BB_\beta = \set {\bigcap \GG: \GG \subseteq \SS_\beta, \GG \text{ is finite} }$.

Lemma 1
Let $V \in \tau_\beta$.

From Inverse of Homeomorphism is Homeomorphism:
 * $\phi^{-1}$ is a homeomorphism

By definition of homeomorphism:
 * $\phi^{-1} \sqbrk V \in \tau_\alpha$

By definition of sub-basis:

We have:

Let $\AA' = \set{\phi \sqbrk B : B \in \BB_\alpha}$

From Lemma 1:
 * $\AA' \subseteq \BB_\beta$

We have:
 * $V = \bigcup \AA'$

By definition, $\SS_\beta$ is a sub-basis for $\tau_\beta$