Describe how to make a topology from a basis, and what (if any) requirements there are on the basis.
Solution
A collection \(\mathcal B\) of subsets of \(X\) is a basis if (i) every \(x \in X\) lies in some \(B \in \mathcal B\), and (ii) whenever \(x \in B_1 \cap B_2\) with \(B_1, B_2 \in \mathcal B\), there is \(B_3 \in \mathcal B\) with \(x \in B_3 \subseteq B_1 \cap B_2\). Declare \(U \subseteq X\) open if for every \(x \in U\) there is \(B \in \mathcal B\) with \(x \in B \subseteq U\), and let \(\mathcal T\) be the collection of open sets.
1. \(\mathcal T\) is a topology on \(X\).
\(\varnothing \in \mathcal T\) vacuously, and \(X \in \mathcal T\) by (i). A union of members of \(\mathcal T\) is in \(\mathcal T\), since a basic set around a point of one member lies in the union. If \(U_1, U_2 \in \mathcal T\) and \(x \in U_1 \cap U_2\), choose \(B_i \in \mathcal B\) with \(x \in B_i \subseteq U_i\); by (ii) there is \(B_3 \in \mathcal B\) with \(x \in B_3 \subseteq B_1 \cap B_2 \subseteq U_1 \cap U_2\). So \(\mathcal T\) is closed under finite intersections.
2. The members of \(\mathcal T\) are exactly the unions of members of \(\mathcal B\).
1. Conversely, if \(U \in \mathcal T\), choose for each \(x \in U\) some \(B_x \in \mathcal B\) with \(x \in B_x \subseteq U\); then \(U = \bigcup_{x \in U} B_x\).
3. Q.E.D.
The topology generated by \(\mathcal B\) is \(\mathcal T\), the collection of unions of members of \(\mathcal B\) (step <1>2); conditions (i) and (ii) are the requirements on \(\mathcal B\), and step 1 shows they make \(\mathcal T\) a topology.