# Topology 2

There is an alternate definition of the basis generated by a topology that is often useful:

Theorem 1 Let ${\mathcal B}$ be a basis for a topology ${\mathcal T}$ on ${X}$. Then ${O\in\mathcal T}$ if and only if

$\displaystyle O=\bigcup_{\alpha\in I}U_\alpha$

where ${\{U_\alpha : \alpha\in I \}\subset\mathcal B}$.

Proof: Let ${O\in\mathcal T}$. Then for each ${x\in O}$, there exists ${U_x\in\mathcal T}$ such that ${x\in U\subset O}$. It follows that ${O=\bigcup_{x\in X}U_x}$. Conversely, if ${O}$ is the union of open sets ${U_\alpha}$, then for each ${x\in O}$, there is a ${\beta}$ such that ${x\in U_\beta}$, and so there is a basis element ${B\in\mathcal B}$ with ${x\in B\subset U_\beta\subset O}$, which implies that ${O\in\mathcal T}$. $\Box$

Bases are convenient for defining topologies whose members cannot easily be generically expressed. For example, given a metric space ${(X,d)}$, the collection of open balls, that is, sets of the form

$\displaystyle B(x,r)=\{y\in X: d(x,y)

is a basis for topology on ${X}$. For clearly

$\displaystyle X=\bigcup_{x\in X}B(x,1),$

and if ${x\in B(y,r)\cap B(z,q)}$ then ${B(x, r-d(x,y))\subset B(y,r)}$ and ${B(x, q-d(x,z))\subset B(z,q)}$, so letting ${r^*=\min\{r-d(x,y), q-d(x,z)\}}$, we have

$\displaystyle x\in B(x, r^*)\subset B(y,r)\cap B(z,q).$

Indeed, the “usual topology” on ${\mathbb R}$, with a set ${U\subset\mathbb R}$ being open iff for each ${x\in U}$ there exists ${\varepsilon_x>0}$ such that ${|x-y|<\varepsilon_x}$ implies ${y\in U}$, is generated by this basis. If ${U}$ is open in ${\mathbb R}$, then

$\displaystyle U=\bigcup_{x\in U} \left\{y\in \mathbb R:|x-y|<\varepsilon_x \right\},$

that is, open sets in ${\mathbb R}$ are unions of open balls.

More generally, there is the notion of a subbasis:

Definition 2 A collection of subsets ${\mathcal S}$ of ${X}$ is a subbasis for a topology on ${X}$ if ${X = \bigcup_{S\in\mathcal S}S}$. The topology generated by ${\mathcal S}$ is the collection of sets of the form

$\displaystyle \bigcup_{T\subset S}\bigcap_{\substack{S\subset\mathcal S\\ S\text{finite}}} T.$

In particular, given a family of topological spaces ${\{(X_\alpha,\mathcal T_\alpha):\alpha\in I\}}$, we define the product topology as that generated by the subbasis

$\displaystyle \mathcal S = \bigcup_{\alpha\in I} \{\pi_\alpha^{-1}(U): U\in\mathcal T_\alpha\},$

where

$\displaystyle \pi_\beta:\bigcup_{\alpha\in I} X_\alpha \rightarrow X_\beta$

is the projection map defined by ${\pi_\beta((x_\alpha)_{\alpha\in I}) = x_\beta}$.

Theorem 3 The box topology is the same as the product topology when ${I}$ is finite and strictly finer than the product topology when ${I}$ is infinite.

Proof: If ${|I|=n}$, then for any ${x\in \prod_{i=1}^n X_i=:X}$ and ${U=\prod_{i=1}^n U_i}$ open in ${X}$ with ${x\in U}$, we have

$\displaystyle U=\prod_{i=1}^n \pi_i^{-1}(U_i),$

so ${U}$ is an element of the basis for the product topology.

If ${I}$ is infinite, then an element ${B}$ of the basis ${\mathcal B}$ for the product topology is of the form

$\displaystyle B = \prod_{i=1}^n \pi_{\beta_i}^{-1}(U_{\beta_i}),$

where ${U_{\beta_i}}$ is open in ${X_{\beta_i}}$, so

$\displaystyle B=\prod_{i\in I} V_\alpha,$

where

$\displaystyle V_{\alpha} = \begin{cases} \pi_{\alpha}^{-1}(U_\alpha),& \alpha\in\{\beta_1,\ldots,\beta_n\}\\ X_\alpha,& \text{ otherwise}. \end{cases}$

If ${(U_\alpha)_{\alpha\in I}}$ is open in ${U_\alpha}$ for each ${\alpha\in I}$, then

$\displaystyle U:=\prod_{\alpha\in i}U_\alpha$

is an element of ${\beta}$ if and only if ${U_\alpha=X_\alpha}$ for all but finitely many ${\alpha}$.

$\Box$

For example, the box topology on ${\mathbb R^n}$ is generated by the open rectangles

$\displaystyle \prod_{i=1}^n (x_i-\varepsilon_i, x_i+\varepsilon_i).$

If ${U=\prod_{i=1}^n (x_i-\varepsilon_i, x_i+\varepsilon_i)}$ then

$\displaystyle U=\prod_{i=1}^n \pi_i^{-1}((x_i-\varepsilon_i, x_i+\varepsilon_i)),$

is also open in the product topology.