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

${\mathcal {B}}:=\left\{\bigcap _{j=1}^{n}f_{\alpha _{j}}^{-1}(U_{j})\mid n\in \mathbb {N} ,\alpha _{1},\ldots ,\alpha _{n}\in A,U_{1},\ldots ,U_{n}\in \tau _{\alpha }\right\}$.

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

${\mathcal {B}}:=\left\{\left\{(x_{\beta })_{\beta \in A}\in X\mid \forall j\in [n]:x_{\alpha _{j}}\in U_{j}\right\}\mid n\in \mathbb {N} _{0},\alpha _{1},\ldots ,\alpha _{n}\in A,U_{1}\in \tau _{\alpha _{1}},\ldots ,U_{n}\in \tau _{\alpha _{n}}\right\}$

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

$\left\{(x_{\beta })_{\beta \in A}\in X\mid \forall j\in [n]:x_{\alpha _{j}}\in U_{j}\right\}=\pi _{\alpha _{1}}(U_{1})\cap \cdots =\pi _{\alpha _{n}}(U_{n})$,

we conclude. $\Box$