Product topology

Note: Very often confused with the Box topology see Product vs box topology for details

$$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$ $$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$ $$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$$

Definition

Given an arbitrary collection of indexed $(X_\alpha,\mathcal{J}_\alpha)_{\alpha\in I}$ topological spaces, we define the product topology as follows:

• Let $X:=\prod_{\alpha\in I}X_\alpha$ be a set imbued with the topology generated by the basis:
• $\mathcal{B}=\left\{\prod_{\alpha\in I}U_\alpha\Big\vert\ \forall\alpha\in I[U_\alpha\in\mathcal{J}_\alpha]\wedge\exists n\in\mathbb{N}[\vert\{U_\alpha\vert U_\alpha\ne X_\alpha\}\vert=n]\right\}$
  • That is to say the basis set contains all the products of open sets where the product has a finite number of elements that are not the entirety of their space.
  • For the sake of contrast, the Box topology has this definition for a basis:
    $\mathcal{B}_\text{box}=\left\{\prod_{\alpha\in I}U_\alpha\Big\vert\ \forall\alpha\in I[U_\alpha\in\mathcal{J}_\alpha]\right\}$ - the product of any collection of open sets
• Note that in the case of a finite number of spaces, say $(X_i,\mathcal{J}_i)_{i=1}^n$ then the topology on $\prod_{i=1}^nX_i$ is generated by the basis:
  • $\mathcal{B}_\text{finite}=\left\{\prod^n_{i=1}U_i\Big\vert\ \forall i\in\{1,2,\ldots,n\}[U_i\in\mathcal{J}_i]\right\}$ (that is to say the box/product topologies agree)

Characteristic property

Here $p_i$ denotes the canonical projection, sometimes $\pi_i$ is used - I avoid using $\pi$ because it is too similar to $\prod$ (at least with my handwriting) - I have seen books using both of these conventions

---

TODO: Finish off

---

$$\begin{xy} \xymatrix{ & \prod_{\beta\in I}X_\beta \ar[d]^{p_i} \\ Y \ar[ur]^f \ar[r]_{f_i} & X_i }\end{xy}$$
(Commutes $\forall \alpha\in I$)