(Lemma) Obtaining a basis from a topology

Lemma (Munkres 13.2): Obtaining a basis for a given topology

Let $(X,\mathcal{T})$ be a topological space and let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of open sets of $X$ such that for each open set $U\subseteq X$ and each $x\in U$, there exists $C\in \mathcal{C}$ such that $x\in C\subseteq U$. Then (1) $\mathcal{C}$ is a basis and (2) $\mathcal{C}$ generates $\mathcal{T}$.

Sketch.

1. Show $\mathcal{C}$ is a basis.
   • (B1) Just take $X\in \mathcal{T}$.
   • (B2) Since $C_1,C_2\in \mathcal{C}$ are open by hypothesis, we have $C_1\cap C_2\in \mathcal{T}$ open as well, so by hypothesis there exists some $C_3$ that satisfies the condition $x\in C_3\subseteq C_1\cap C_2$.
2. Show that the topology generated by $\mathcal{C}$ and the topology $\mathcal{T}$ are equivalent using double containment.
   • $\mathcal{T}\subseteq {\mathcal{T}}_{\mathcal{C}}$. By hypothesis, $U\in \mathcal{T}$ and $x\in U⟹$ there exists $C\in \mathcal{C}$ such that $x\in C\subseteq U$, so $U\in {\mathcal{T}}_{\mathcal{C}}$.
   • ${\mathcal{T}}_{\mathcal{C}}\subseteq \mathcal{T}$. Since every open set in a topology is equal to a union of basis elements in a topology, $W\in {\mathcal{T}}_{\mathcal{C}}⟹$ $W$ is some union of elements $C\in \mathcal{C}$, and $\mathcal{C}\subseteq \mathcal{T}⟹$ unions of $C\in \mathcal{C}$ are back in $\mathcal{T}⟹W\in \mathcal{T}$ as well.