> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. A family of [[Open Set|open sets]] $\cb \subset \ct$ is a **base** for $X$ if it contains a [[Neighbourhood Base|neighbourhood base]] for every $x \in X$. > [!theorem] > > Let $(X, \topo)$ be a topological space and $\ce \subset \ct$. $\ce$ is a base if and only if every non-empty $U \in \ct$ is a union of members of $\ce$. > > *Proof*. Suppose that $\ce$ is a base, and let $U \in \ct$. Then for any $x \in U$, $\ce$ contains a neighbourhood base of $x$, where we can find $V_x \in \ce$ such that $x \in V_x$ and $V_x \subset U$. Then > $ > U = \bigcup_{x \in U}V_x \quad x \in V_x, V_x \subset U \quad \forall x \in U > $ > > Now suppose that every non-empty $U \in \topo$ is a union of members of $\ce$. Let $x \in X$ and $\ce_x = \bracs{V \in \ce: V \ni x}$, then $\ce_x$ is non-empty since $X \in \ce_x$. Let $U \in \topo$, $U \ni x$, then > $ > U = \bigcup_{i \in I}V_i \quad V_i \in \ce > $ > and as $x \in U$, $\exists i \in I: x \in V_i$. So we have found $V_i \in \ce_x$ such that $V_i \subset U$. Therefore $\ce_x$ is a neighbourhood base at $x$, and $\ce$ is a base for $\topo$. > [!theorem] > > Let $X$ be a non-empty set and $\ce \subset \pow{X}$. Then $\ce$ is a base for *some* topology $\topo$ on $X$ if and only if > - Each $c \in X$ is contained in some $V \in \ce$. > - If $U, V \in \ce$ and $x \in U \cap V$, then there exists $W \in \ce$ with $x \in W \subset (U \cap V)$. > > *Proof*. Suppose that $\ce$ is a base for some topology $\topo$ on $X$, then it contains a neighbourhood base $\ce_x$ for each $x \in X$. So for any $x \in X$, there exists $V_x \in \ce_x$ such that $x \in V_x$. Let $U, V \in \ce$, then $U \cap V \in \topo$ is another open set. If $U \cap V \ne \emptyset$, then for any $x \in U \cap V$ there exists $V_x$ such that $x \in V_x \subset U \cap V$. > > Now suppose that the above conditions hold. Let > $ > \topo = \bracs{U \subset X: \forall x \in U, \exists V \in \ce: x \in V \subset U} > $ > Then $\emptyset, X \in \topo$ and $\ce \subset \topo$. > > Let $\bracs{U_i}_{i \in I} \subset \topo$ be a collection of sets in $\topo$, and $U = \bigcup_{i \in I}U_i$. Then for any $x \in X$, there exists $i \in I$ such that $x \in U_i$, and $\exists V \in \ce$ such that $x \in V \subset U_i \subset U$. Therefore $U \in \topo$, and $\topo$ is closed under arbitrary unions. > > Let $U, V \in \topo$ be two sets in $\topo$, and $x \in U \cap V$. Then there exists $U'$ and $V'$ such that $x \in U' \subset U$, $x \in V' \subset V$, and there exists $W$ such that $x \in W \subset U' \cap V' \subset U \cap V$. So $U \cap V \in \topo$ as well, and $\topo$ is closed under finite intersections. > > Now let $U \in \topo$, then for any $x \in U$ there exists $V_x \in \ce$ such that $x \in V_x \subset U$, and > $ > U = \bigcup_{x \in U}V_x \quad x \in V_x \subset U \quad \forall x \in U > $ > we have $\ce$ being a base for $\topo$.