The topology generated by any subset \mathcal{S} \subseteq \{\varnothing, X\} (including by the empty set \mathcal{S} := \varnothing) is equal to the trivial topology \{\varnothing, X\}. If \tau is a topology on X and \mathcal{B} is a basis for \tau then the topology generated by \mathcal{B} is \tau. Thus any basis \mathcal{B} for a topology \tau is also a subbasis for \tau. If \mathcal{S} is any subset of \tau then the topology generated by \mathcal{S} will be a subset of \tau. The usual topology on the
real numbers \R has a subbase consisting of all
semi-infinite open intervals either of the form (-\infty, a) or (b, \infty), where a and b are real numbers. Together, these generate the usual topology, since the intersections (a,b) = (-\infty, b) \cap (a, \infty) for a \leq b generate the usual topology. A second subbase is formed by taking the subfamily where a and b are
rational. The second subbase generates the usual topology as well, since the open intervals (a, b) with a, b rational, are a basis for the usual Euclidean topology. The subbase consisting of all semi-infinite open intervals of the form (-\infty, a) alone, where a is a real number, does not generate the usual topology. The resulting topology does not satisfy the
T1 separation axiom, since if a every
open set containing b also contains a. The
initial topology on X defined by a family of functions f_i : X \to Y_i, where each Y_i has a topology, is the coarsest topology on X such that each f_i is
continuous. Because continuity can be defined in terms of the
inverse images of open sets, this means that the initial topology on X is given by taking all f_i^{-1}(U), where U ranges over all open subsets of Y_i, as a subbasis. Two important special cases of the initial topology are the
product topology, where the family of functions is the set of projections from the product to each factor, and the
subspace topology, where the family consists of just one function, the
inclusion map. The
compact-open topology on the space of continuous functions from X to Y has for a subbase the set of functions V(K,U) = \{f : X \to Y \mid f(K) \subseteq U\} where K \subseteq X is
compact and U is an open subset of Y. Suppose that (X, \tau) is a
Hausdorff topological space with X containing two or more elements (for example, X = \R with the
Euclidean topology). Let Y \in \tau be any non-empty subset of (X, \tau) (for example, Y could be a non-empty bounded open interval in \R) and let \nu denote the
subspace topology on Y that Y inherits from (X, \tau) (so \nu \subseteq \tau). Then the topology generated by \nu
on X is equal to the union \{X\} \cup \nu (see the footnote for an explanation), where \{X\} \cup \nu \subseteq \tau (since (X, \tau) is Hausdorff, equality will hold if and only if Y = X). Note that if Y is a
proper subset of X, then \{X\} \cup \nu is the smallest topology
on X containing \nu yet \nu does not cover X (that is, the union \bigcup_{V \in \nu} V = Y is a proper subset of X). ==Results using subbases==