# General Topology/Bases

Definition (subbasis of a topology):

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

Definition (basis of a topology):

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

${\displaystyle U=\bigcup _{\alpha \in A}B_{\alpha }}$, where ${\displaystyle \forall \alpha \in A:B_{\alpha }\in {\mathcal {B}}}$.

Proposition (basis criterion):

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

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

${\displaystyle U_{1}\cap U_{2}=\bigcup _{\alpha \in A}U_{\alpha }}$, where ${\displaystyle \forall \alpha \in A:U_{\alpha }\in {\mathcal {B}}}$.

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

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

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

${\displaystyle U=\bigcup _{x\in U}U_{x}}$. ${\displaystyle \Box }$

Proposition (basis from subbasis via finite intersections):

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

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

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

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

Proposition (basis of the initial topology):

Let ${\displaystyle X}$ be a topological space, let ${\displaystyle (Y_{\alpha })_{\alpha \in A}}$ be topological spaces, and let ${\displaystyle f_{\alpha }:X\to Y_{\alpha }}$ be functions. If we denote the topology of each ${\displaystyle Y_{\alpha }}$ by ${\displaystyle \tau _{\alpha }}$, then a basis for the initial topology on ${\displaystyle X}$ is given by

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

Proof: First we note that ${\displaystyle {\mathcal {B}}}$ is contained within the initial topology. Further, the initial topology is also the smallest topology that contains ${\displaystyle {\mathcal {B}}}$, since any topology that contains ${\displaystyle {\mathcal {B}}}$ contains all the individual initial topologies ${\displaystyle f_{\alpha }^{-1}(\tau _{\alpha })}$. Then, using the characterisation of the generated topology we gave, we note that we may write a set ${\displaystyle U}$ which is in the topology generated by the individual topologies ${\displaystyle f_{\alpha }^{-1}(\tau _{\alpha })}$ as ${\displaystyle \bigcup _{\beta \in B}C_{\beta }}$, ${\displaystyle C_{\beta }=D_{\beta ,1}\cap \cdot D_{\beta ,n_{\beta }}}$, ${\displaystyle D_{\beta ,j}f_{\alpha _{\beta ,j}}^{-1}(\tau _{\alpha _{\beta ,j}})}$. ${\displaystyle \Box }$

Proposition (basis of the product topology):

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

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

constitutes a basis for the product topology on ${\displaystyle X}$.

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

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

we conclude. ${\displaystyle \Box }$