General Topology/Bases

Definition (subbasis of a topology):

Suppose that $\tau$ is the topology on $X$ generated by a set ${\mathcal {B}}$ . Then ${\mathcal {B}}$ is called a subbasis of the topology $\tau$ .

Definition (basis of a topology):

Let $X$ be a topological space, where $\tau$ is its topology. A basis of $\tau$ is a set ${\mathcal {B}}\subseteq {\mathcal {P}}(X)$ so that every $U\in \tau$ may be written as a union of elements of ${\mathcal {B}}$ , that is,

U=\bigcup _{\alpha \in A}B_{\alpha }

, where

\forall \alpha \in A:B_{\alpha }\in {\mathcal {B}}

.

Proposition (basis criterion):

Let ${\mathcal {B}}$ be a set of subsets of a set $X$ . ${\mathcal {B}}$ forms the basis of the topology $\tau$ generated by it if and only if for all $U_{1},U_{2}\in {\mathcal {B}}$ and $x\in U_{1}\cap U_{2}$ there exists $U_{3}\in {\mathcal {B}}$ such that $x\in U_{3}\subseteq U_{1}\cap U_{2}$ .

Proof: Suppose first that ${\mathcal {B}}$ does form a basis of the topology $\tau$ generated by it. Then whenever $U_{1},U_{2}\in {\mathcal {B}}$ , the set $U_{1}\cap U_{2}$ is open, so that we may write it as a union

U_{1}\cap U_{2}=\bigcup _{\alpha \in A}U_{\alpha }

, where

\forall \alpha \in A:U_{\alpha }\in {\mathcal {B}}

.

In particular, if $x\in U_{1}\cap U_{2}$ , we find a $U_{\alpha }\in {\mathcal {B}}$ such that $x\in U_{\alpha }$ . Upon setting $U_{3}:=U_{\alpha }$ , we obtain $x\in U_{3}\subseteq U_{1}\cap U_{2}$ . Suppose conversely that ${\mathcal {B}}\subseteq {\mathcal {P}}(X)$ satisfies the given condition. By the characterisation of the topology generated by a set, for every $U\in \tau$ we may write

U=\bigcup _{\alpha \in A}\bigcap _{j=1}^{n_{\alpha }}B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }}

,

where $A$ is an index set and $B_{\alpha ,j}\in {\mathcal {B}}$ for all $\alpha$ and $j$ . Let $\alpha$ be fixed, and let $x\in B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }}$ be arbitrary. Suppose that for $k<n_{\alpha }$ , we found a set $U_{x,k}\subseteq B_{\alpha ,1}\cap \dots \cap B_{\alpha ,k}$ so that $U_{x,k}\in {\mathcal {B}}$ and $x\in U_{x,k}$ . Then by the condition, we pick $U_{x,k+1}\subseteq U_{x,k}\cap B_{\alpha ,k+1}$ so that $U_{x,k+1}\in {\mathcal {B}}$ and $x\in U_{x,k+1}$ , so that finally we end up with a set $U_{x,n_{\alpha }}$ that is in ${\mathcal {B}}$ , in $B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }}$ and contains $x$ . For each $x$ , choose an $\alpha$ so that $x\in B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }}$ and then set $U_{x}$ to be the corresponding $U_{x,n_{\alpha }}$ as constructed above. Then

U=\bigcup _{x\in U}U_{x}

.

\Box

Proposition (basis from subbasis via finite intersections):

Let $X$ be a set and let ${\mathcal {B}}\subseteq {\mathcal {P}}(X)$ . Let $\tau$ be the topology generated by ${\mathcal {B}}$ (ie. ${\mathcal {B}}$ is a subbasis of $\tau$ ) and let

{\mathcal {B}}':=\{B_{1}\cap \cdots \cap B_{n}|n\in \mathbb {N} ,B_{1},\ldots ,B_{n}\in {\mathcal {B}}\}

.

Then ${\mathcal {B}}'$ is a basis for the topology $\tau$ .

Proof: Since ${\mathcal {B}}\subseteq {\mathcal {B}}'$ , clearly the topology generated by ${\mathcal {B}}'$ is a superset of $\tau$ . On the other hand, since $\tau$ is closed under finite intersections, all elements of ${\mathcal {B}}'$ are contained in $\tau$ , so that ${\mathcal {B}}'$ generates the same topology as ${\mathcal {B}}$ . Finally, by the basis criterion, ${\mathcal {B}}'$ is a basis of the topology $\tau$ . $\Box$

Proposition (basis of the initial topology):

Let $X$ be a topological space, let $(Y_{\alpha })_{\alpha \in A}$ be topological spaces, and let $f_{\alpha }:X\to Y_{\alpha }$ be functions. If we denote the topology of each $Y_{\alpha }$ by $\tau _{\alpha }$ , then a basis for the initial topology on $X$ 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 ${\mathcal {B}}$ is contained within the initial topology. Further, the initial topology is also the smallest topology that contains ${\mathcal {B}}$ , since any topology that contains ${\mathcal {B}}$ contains all the individual initial topologies $f_{\alpha }^{-1}(\tau _{\alpha })$ . Then, using the characterisation of the generated topology we gave, we note that we may write a set $U$ which is in the topology generated by the individual topologies $f_{\alpha }^{-1}(\tau _{\alpha })$ as $\bigcup _{\beta \in B}C_{\beta }$ , $C_{\beta }=D_{\beta ,1}\cap \cdot D_{\beta ,n_{\beta }}$ , $D_{\beta ,j}f_{\alpha _{\beta ,j}}^{-1}(\tau _{\alpha _{\beta ,j}})$ . $\Box$

Proposition (basis of the product topology):

Let $(X_{\alpha })_{\alpha \in A}$ be a family of topological spaces, and suppose that $\tau _{\alpha }$ is the topology of $X_{\alpha }$ for each $\alpha \in A$ . Set $X:=\prod _{\alpha \in A}X_{\alpha }$ . 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 $X$ .

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. $\Box$

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