Basis

We will verify it's a topology from the definition

$$ \mathrm{\varnothing <!--\; \varnothing \; -->}$$ is in $$ G_{\mathcal{B}}$$ because it vacuously satisfies the condition. $$ X$$ is in $$ G_{\mathcal{B}}$$ since given any point $$ x\in <!--\; \in \; -->X$$, we have a $$ B\in <!--\; \in \; -->\mathcal{B}$$ such that $$ x\in <!--\; \in \; -->B\subseteq <!--\; \subseteq \; -->X$$ by the definition of a basis

Now suppose we have an arbitrary union of elements from $$ G_{\mathcal{B}}$$, say $$ S\subseteq <!--\; \subseteq \; -->G_{\mathcal{B}},M=\underset{U\in <!--\; \in \; -->S}{\bigcup <!--\; \bigcup \; -->}U$$. We need to verify that $$ M\in <!--\; \in \; -->G_{\mathcal{B}}$$, so let $$ x\in <!--\; \in \; -->M$$, therefore $$ x\in <!--\; \in \; -->\stackrel{¯<!--\; ¯\; -->}{U}$$ for some $$ \stackrel{¯<!--\; ¯\; -->}{U}\in <!--\; \in \; -->S$$ and since $$ \stackrel{¯<!--\; ¯\; -->}{U}$$ was assumed to be in $$ G_{\mathcal{B}}$$ we know that there is some $$ B\in <!--\; \in \; -->\mathcal{B}$$ such that $$ x\in <!--\; \in \; -->B\subseteq <!--\; \subseteq \; -->\stackrel{¯<!--\; ¯\; -->}{U}$$ but also $$ \stackrel{¯<!--\; ¯\; -->}{U}\subseteq <!--\; \subseteq \; -->M$$ since $$ M$$ includes $$ \stackrel{¯<!--\; ¯\; -->}{U}$$ in the union so then $$ x\in <!--\; \in \; -->B\subseteq <!--\; \subseteq \; -->M$$ meaning by the definition that $$ M\in <!--\; \in \; -->G_{\mathcal{B}}$$

Now if we are given a finite intersection of elements from $$ G_{\mathcal{B}}$$ we will prove by induction that it is also an element of $$ G_{\mathcal{B}}$$

• base case

if $$ n=1$$ then the union of one element from $$ G_{\mathcal{B}}$$ (namely just itself) is also an element of $$ G_{\mathcal{B}}$$ trivially.

Before moving to the inductive step, notice that it also holds for $$ n=2$$, because given $$ U_1,U_2\in <!--\; \in \; -->G_{\mathcal{B}}$$, we can see that for any $$ x\in <!--\; \in \; -->U_1\cap <!--\; \cap \; -->U_2$$, we know that we have a basis elements $$ B_1,B_2\in <!--\; \in \; -->\mathcal{B}$$ such that $$ x\in <!--\; \in \; -->B_1\subseteq <!--\; \subseteq \; -->U_1$$ and $$ x\in <!--\; \in \; -->B_2\subseteq <!--\; \subseteq \; -->U_2$$ and thus $$ x$$ also belongs to the intersection of $$ B_1$$ and $$ B_2$$, so by the definition of a basis, we get a $$ B_3\subseteq <!--\; \subseteq \; -->B_1\cap <!--\; \cap \; -->B_2$$ that contains $$ x$$ since $$ B_1\cap <!--\; \cap \; -->B_2\subseteq <!--\; \subseteq \; -->U_1\cap <!--\; \cap \; -->U_2$$, we have $$ x\in <!--\; \in \; -->B_3\subseteq <!--\; \subseteq \; -->U_1\cap <!--\; \cap \; -->U_2$$, which means that $$ U_1\cap <!--\; \cap \; -->U_2\in <!--\; \in \; -->G_{\mathcal{B}}$$

• inductive step

Let $$ k\in <!--\; \in \; -->{\mathbb{N}}_1$$ and assume that the intersection of $$ k$$ elements from $$ G_{\mathcal{B}}$$ is also in $$ G_{\mathcal{B}}$$ now we'll show it's true for $$ k+1$$. So consider the intersection $$ U_1\cap <!--\; \cap \; -->\dots <!--\; \dots \; -->\cap <!--\; \cap \; -->U_{k+1}$$, then we can re-write it as $$ (U_1\cap <!--\; \cap \; -->\dots <!--\; \dots \; -->\cap <!--\; \cap \; -->U_k)\cap <!--\; \cap \; -->U_{k+1}$$ so by our inductive hypothesis $$ (U_1\cap <!--\; \cap \; -->\dots <!--\; \dots \; -->\cap <!--\; \cap \; -->U_k)\in <!--\; \in \; -->G_{\mathcal{B}}$$ and therefore this $$ k+1$$ union is a union of two elements already in $$ G_{\mathcal{B}}$$ so by our $$ n=2$$ argument $$ U_1\cap <!--\; \cap \; -->\dots <!--\; \dots \; -->\cap <!--\; \cap \; -->U_{k+1}\in <!--\; \in \; -->G_{\mathcal{B}}$$, as needed.

Symbolically we can write this collection of unions as $$ \mathcal{U}=\{\underset{B\in <!--\; \in \; -->\mathcal{S}}{\bigcup <!--\; \bigcup \; -->}B:\mathcal{S}\subseteq <!--\; \subseteq \; -->\mathcal{C}\}$$

Suppose we have a subset of elements from $$ \mathcal{B}$$ called $$ \mathcal{S}$$, then for each $$ B\in <!--\; \in \; -->\mathcal{S}$$ it is open in $$ X$$ because it is a basis element, and therefore by the definition of a topology, as $$ \underset{B\subseteq <!--\; \subseteq \; -->\mathcal{S}}{\bigcup <!--\; \bigcup \; -->}B$$ is an arbitrary union of open sets it is also open in $$ X$$, so we've shown that $$ \mathcal{U}\subseteq <!--\; \subseteq \; -->{\mathcal{T}}_{\mathcal{B}}$$

We'll show the converse now, so suppose that $$ U\in <!--\; \in \; -->{\mathcal{T}}_{\mathcal{B}}$$, we want to prove that it's equal to an arbitrary union of elements of $$ \mathcal{B}$$.

Recall the definition of a topology generated by a basis so since $$ U\in <!--\; \in \; -->{\mathcal{T}}_{\mathcal{B}}$$ then for each $$ x\in <!--\; \in \; -->U$$ we have some $$ B_x\in <!--\; \in \; -->\mathcal{B}$$ with $$ x\in <!--\; \in \; -->B_x\subseteq <!--\; \subseteq \; -->U$$, therefore $$ U=\underset{x\in <!--\; \in \; -->U}{\bigcup <!--\; \bigcup \; -->}{B}_x$$, and as a union of of elements of $$ \mathcal{B}$$ thus $$ U$$ is an element of $$ \mathcal{U}$$ so $$ {\mathcal{T}}_{\mathcal{B}}\subseteq <!--\; \subseteq \; -->\mathcal{U}$$.

We'll first show that $$ \mathcal{C}$$ is a basis, so let $$ x\in <!--\; \in \; -->X$$, but $$ X$$ is open and so we can use $$ X$$ as the element of $$ \mathcal{C}$$ satisfying $$ x\in <!--\; \in \; -->X\subseteq <!--\; \subseteq \; -->X$$.

Now let $$ C_1,C_2$$ be elements of $$ \mathcal{C}$$ and now consider $$ C_1\cap <!--\; \cap \; -->C_2$$, since it's a finite intersection, then we know it's also open in $$ X$$ from the definition of a topology.

Since $$ C_1\cap <!--\; \cap \; -->C_2$$ is open, let $$ x\in <!--\; \in \; -->C_1\cap <!--\; \cap \; -->C_2$$, then by the definition of $$ \mathcal{C}$$, there is some $$ C\in <!--\; \in \; -->\mathcal{C}$$ such that $$ x\in <!--\; \in \; -->C\subseteq <!--\; \subseteq \; -->C_1\cap <!--\; \cap \; -->C_2$$, so we can take $$ B_3=C$$ in the definition of basis, to see that the second condition is satisfied, and thus we know that $$ \mathcal{C}$$ is a basis

Now we want to show that $$ {\mathcal{T}}_{\mathcal{C}}=\mathcal{T}$$. So suppose $$ U\in <!--\; \in \; -->{\mathcal{T}}_{\mathcal{C}}$$, then it can be written as a union as elements from $$ \mathcal{C}$$, but since $$ \mathcal{C}$$ was defined to consist of open sets of $$ X$$, then this union is also open with respect to $$ X$$ by the definition of a topology.

Now suppose that $$ U\in <!--\; \in \; -->\mathcal{T}$$ and consider some $$ x\in <!--\; \in \; -->U$$ then by the definition of $$ \mathcal{C}$$, there is some $$ C\in <!--\; \in \; -->\mathcal{C}$$ such that $$ x\in <!--\; \in \; -->C\subseteq <!--\; \subseteq \; -->U$$, looking back at the definition for a topology generated by a basis, we can see this means $$ U\in <!--\; \in \; -->{\mathcal{T}}_{\mathcal{C}}$$, as needed.