# A Basis for Arbitrary Topological Products of Topological Spaces

Recall from the Arbitrary Topological Products of Topological Spaces page that if $\{ X_i \}_{i \in I}$ is an arbitrary collection of topological spaces and $\displaystyle{\prod_{i \in I} X_i}$ is the Cartesian product of these topological spaces whose elements are sequences $(x_i)_{i \in I}$ where $x_i$ is the $i^{\mathrm{th}}$ component of $(x_i)_{i \in I}$, then the product topology $\tau$ on $\displaystyle{\prod_{i \in I} X_i}$ is the initial topology induced by the projection maps $\displaystyle{p_i : \prod_{i \in I} \to X_i}$ and the resulting topological product is the topological space $\left ( \prod_{i \in I} X_i, \tau \right )$.

We also noted that a subbasis for the product topology is:

(1)In other words, the elements in the subbasis $\mathcal S$ are the subsets $\displaystyle{U \subseteq \prod_{i \in I} X_i}$ whose inverse image under some projection $p_i$ is equal to an open set $V$ in $X_i$.

We will now give an explicit form for a basis of the product topology on an arbitrary topological product.

Theorem 1: Let $\{ X_i \}_{i \in I}$ be an arbitrary collection of topological spaces and let $\displaystyle{\prod_{i \in I} X_i}$ be the Cartesian product of these spaces. Then a basis for the product topology is $\displaystyle{\mathcal B = \left \{ U = \prod_{i \in I} U_i : U_i \subseteq X_i \: \mathrm{is \: open \: in \:} X_i \: \mathrm{and} U_i = X_i \: \mathrm{for \: all \: but \: finitely \: many} \: i \right \}}$. |

**Proof:**Take $\mathcal S$ as above as a subbasis for the product topology $\tau$ on $\displaystyle{\prod_{i \in I} X_i}$. Then the collection of all finite intersections of elements from $\mathcal S$ form a basis of $\tau$. Consider an arbitrary finite intersection $U$ of elements from $\mathcal S$:

- Since $\mathcal S$ is a subbasis of $\tau$ we have that $U$ is an open set in $\displaystyle{\prod_{i \in I} X_i}$. But this means that for some $i \in I$, $p_i^{-1}(U) = V$ is an open set in $X_i$ for some $i \in I$. This happens if $U_{i, 1}, U_{i, 2}, ..., U_{i, n}$ are all open in $X_i$. So $\displaystyle{U = \prod_{i \in I} U_i}$ if $U_i = X_i$ for all but finitely many $i$. This shows that $\mathcal B$ is the basis obtained from $\mathcal S$. $\blacksquare$