Bases and subbases are economical ways to specify a topology. They let one prove continuity and construct spaces without listing every open set.
Bases for a topology
A basis gives small open pieces from which all open sets are assembled.
Definition. A family \(\mathcal B\) of subsets of \(X\) is a basis if for every \(x\in X\) some \(B\in\mathcal B\) contains \(x\), and whenever \(x\in B_1\cap B_2\) with \(B_1,B_2\in\mathcal B\), there is \(B_3\in\mathcal B\) such that \(x\in B_3\subseteq B_1\cap B_2\).
Proposition. If \(\mathcal B\) is a basis, the unions of members of \(\mathcal B\) form a topology on \(X\).
Proof. The empty union is \(\varnothing\), and the covering condition gives \(X=\bigcup\mathcal B\). Arbitrary unions of unions of basis elements are again unions of basis elements. For finite intersections, it is enough to intersect two basic unions; if \(x\) lies in the intersection, choose \(B_1,B_2\in\mathcal B\) containing \(x\) inside the two unions, then refine to \(B_3\subseteq B_1\cap B_2\).
Local tests with bases
Once a basis is fixed, open-set statements can be checked on basic neighborhoods.
Proposition. Let \(\mathcal B\) be a basis for \(X\). A subset \(U\subseteq X\) is open if and only if for every \(x\in U\) there exists \(B\in\mathcal B\) with \(x\in B\subseteq U\).
Proof. If \(U\) is open, it is a union of basis elements, so each \(x\in U\) lies in some basic \(B\subseteq U\). Conversely, if every point of \(U\) has such a basic neighborhood, then \(U\) is the union of all these chosen basis elements and is open.
Corollary. A map \(f:X\to Y\) is continuous if it suffices to check that \(f^{-1}(B)\) is open for every member \(B\) of a basis for \(Y\).
Proof. Every open set in \(Y\) is a union of basic open sets. Inverse images preserve arbitrary unions, so the inverse image of an arbitrary open set is a union of open sets in \(X\).
Subbases and finite intersections
A subbasis specifies even less data: finite intersections first make a basis, then arbitrary unions make the topology.
Definition. A subbasis \(\mathcal S\) for a topology on \(X\) is a family whose union covers \(X\). The topology generated by \(\mathcal S\) has as a basis all finite intersections of members of \(\mathcal S\), with \(X\) allowed as the empty finite intersection.
Proposition. The topology generated by \(\mathcal S\) is the smallest topology on \(X\) containing \(\mathcal S\).
Proof. Every topology containing \(\mathcal S\) must contain finite intersections of elements of \(\mathcal S\) and then arbitrary unions of those finite intersections. The topology constructed from these finite intersections therefore lies inside every topology containing \(\mathcal S\), while itself containing \(\mathcal S\).
Alexander subbase theorem
Subbases are not merely notation; they reduce compactness checks to a smaller family of covers.
Theorem. A space \(X\) with subbasis \(\mathcal S\) is compact if every cover of \(X\) by members of \(\mathcal S\) has a finite subcover.
Proof. Assume \(X\) is not compact. By the maximal principle, choose an open cover \(\mathcal U\) with no finite subcover that is maximal among such covers. If a subbasic set \(S\in\mathcal S\) is not in \(\mathcal U\), maximality makes \(\mathcal U\cup\{S\}\) finitely cover \(X\), so finitely many members of \(\mathcal U\) cover \(X\setminus S\). If \(\mathcal U\) contained enough subbasic members to cover \(X\), the hypothesis would give a finite subcover, impossible. Hence some \(x\in X\) is missed by all subbasic members of \(\mathcal U\). But \(\mathcal U\) covers \(X\), so choose \(U\in\mathcal U\) with \(x\in U\), and then a basic finite intersection \(S_1\cap\cdots\cap S_n\) with \(x\) inside it and contained in \(U\). Each \(S_i\) is not in \(\mathcal U\), so finitely many sets of \(\mathcal U\) cover \(X\setminus S_i\). Together with \(U\), these finitely cover \(X\), a contradiction.
Remark. The theorem is the engine behind many proofs of Tychonoff's theorem, because product topologies have very natural subbases.