General Topology/Bases

From Wikibooks, open books for an open world
Jump to navigation Jump to search

Definition (subbasis of a topology):

Suppose that is the topology on generated by a set . Then is called a subbasis of the topology .

Definition (basis of a topology):

Let be a topological space, where is its topology. A basis of is a set so that every may be written as a union of elements of , that is,

, where .

Proposition (basis criterion):

Let be a set of subsets of a set . forms the basis of the topology generated by it if and only if for all and there exists such that .

Proof: Suppose first that does form a basis of the topology generated by it. Then whenever , the set is open, so that we may write it as a union

, where .

In particular, if , we find a such that . Upon setting , we obtain . Suppose conversely that satisfies the given condition. By the characterisation of the topology generated by a set, for every we may write

,

where is an index set and for all and . Let be fixed, and let be arbitrary. Suppose that for , we found a set so that and . Then by the condition, we pick so that and , so that finally we end up with a set that is in , in and contains . For each , choose an so that and then set to be the corresponding as constructed above. Then

.

Proposition (basis from subbasis via finite intersections):

Let be a set and let . Let be the topology generated by (ie. is a subbasis of ) and let

.

Then is a basis for the topology .

Proof: Since , clearly the topology generated by is a superset of . On the other hand, since is closed under finite intersections, all elements of are contained in , so that generates the same topology as . Finally, by the basis criterion, is a basis of the topology .

Proposition (basis of the initial topology):

Let be a topological space, let be topological spaces, and let be functions. If we denote the topology of each by , then a basis for the initial topology on is given by

Failed to parse (unknown function "\middle"): {\displaystyle \mathcal B := \left\{ \bigcap_{j=1}^n f_{\alpha_j}^{-1}(U_j) \middle| n \in \mathbb N, \alpha_1, \ldots, \alpha_n \in A, U_1, \ldots, U_n \in \tau_\alpha \right\}} .

Proof: First we note that is contained within the initial topology. Further, the initial topology is also the smallest topology that contains , since any topology that contains contains all the individual initial topologies . Then, using the characterisation of the generated topology we gave, we note that we may write a set which is in the topology generated by the individual topologies as , , .

Proposition (basis of the product topology):

Let be a family of topological spaces, and suppose that is the topology of for each . Set . Then the set

Failed to parse (unknown function "\middle"): {\displaystyle \mathcal B := \left\{ \left\{ (x_\beta)_{\beta \in A} \in X \middle| \forall j \in [n]: x_{\alpha_j} \in U_j \right\} \middle| n \in \mathbb N_0, \alpha_1, \ldots, \alpha_n \in A, U_1 \in \tau_{\alpha_1}, \ldots, U_n \in \tau_{\alpha_n} \right\}}

constitutes a basis for the product topology on .

Proof: By inspecting the form of the canonical basis of the initial topology, and noting that

Failed to parse (unknown function "\middle"): {\displaystyle \left\{ (x_\beta)_{\beta \in A} \in X \middle| \forall j \in [n]: x_{\alpha_j} \in U_j \right\} = \pi_{\alpha_1}(U_1) \cap \cdots = \pi_{\alpha_n}(U_n)} ,

we conclude.