Exercise 13.5

Proof. Let $T_A$ be the topology generated by $A$, and $T_I$ be the intersection of all topologies that contain $A$.

$T_I\subseteq T_A$. This follows from the fact that $T_A\supseteq A$, and so is one of the topologies that is intersected over in the construction of $T_I$.

$T_A\subseteq T_I$. Let $U\in T_A$; by Lemma 13.1, $U={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha }{A}_{\alpha }$ for some collection ${\{A_{\alpha }\}}_{\alpha }\subseteq A$. But $U={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha }{A}_{\alpha }\in T_I$ since each $A_{\alpha }\in T_I$.

Now let $A$ be a subbasis. The proof that $T_I\subseteq T_A$ is identical; it remains to show $T_A\subseteq T_I$. Let $U\in T_A$; by definition of the topology generated by $A$, $U$ is the union of a finite intersection of elements ${\{A_{\alpha }\}}_{\alpha }\subseteq A$. But then $U\in T_I$ since each $A_{\alpha }\in T_I$. □

Proof. Consider $U\in T$ and any $T_c\in C$ so that $A\subset T_c$. Then, since $A$ generates $T$, it follows from Lemma 13.1 that $U$ is the union of elements of $A$. Clearly then each of these elements of $A$ is in $T_c$ since $A\subset T_c$ so that their union is as well since $T_c$ is a topology. Hence $U\in T_c$ so that $T\subset T_c$ since $U$ was arbitrary. Hence $T$ is contained in all elements of $C$ so that $T\subset \bigcap <!--\; FUNCTION\; APPLICATION\; -->C$. Also, clearly $T$ is a topology that contains $A$ so that $T\in C$. Clearly then $\bigcap <!--\; FUNCTION\; APPLICATION\; -->C\subset T$ so that $T=\bigcap <!--\; FUNCTION\; APPLICATION\; -->C$ as desired. □

Proof. Let $B$ be the set of all finite intersections of elements of $A$, which we know is a basis for $T$ by the proof after the definition of a subbasis. We show that $B\subset T_c$ for all $T_c\in C$. So consider any set $B\in B$ so that $B$ is the finite intersection of elements of $A$. Also consider any $T_c\in C$ so that each of these elements is in $T_c$ since $A\subset T_c$. Since $T_c$ is a topology, clearly, the finite intersection of these elements, i.e. $B$, is in $T_c$. Hence $B\subset T_c$ since $B$ was arbitrary.

It then follows from what was shown before that $T=\bigcap <!--\; FUNCTION\; APPLICATION\; -->C$ since $T$ is the topology generated by the basis $B$ and $B$ is contained in each topology in $C$. □