[proofplan]
We show that the product [topology](/page/Topology) and the box topology on a finite product $\prod_{i=1}^n X_i$ have the same basis, hence coincide. The product topology basis consists of products $\prod_{i=1}^n V_i$ with each $V_i$ open and all but finitely many equal to $X_i$. The box topology basis consists of products $\prod_{i=1}^n V_i$ with each $V_i$ open and no restriction on how many factors differ from $X_i$. When the index set is finite, the finiteness constraint in the product topology is automatically satisfied, so the two bases are identical.
[/proofplan]
[step:Recall the two bases]
The product [topology](/page/Topology) on $\prod_{i=1}^n X_i$ has as a basis the collection
\begin{align*}
\mathcal{B}_{\mathrm{prod}} := \Bigl\{ \prod_{i=1}^n V_i : V_i \in \tau_i \text{ for each } i, \text{ and } V_i = X_i \text{ for all but finitely many } i \Bigr\},
\end{align*}
where $\tau_i$ denotes the topology on $X_i$. The box topology on $\prod_{i=1}^n X_i$ has as a basis the collection
\begin{align*}
\mathcal{B}_{\mathrm{box}} := \Bigl\{ \prod_{i=1}^n V_i : V_i \in \tau_i \text{ for each } i \Bigr\}.
\end{align*}
The only difference is that $\mathcal{B}_{\mathrm{prod}}$ imposes the constraint that all but finitely many factors equal the full space $X_i$, while $\mathcal{B}_{\mathrm{box}}$ imposes no such constraint.
[/step]
[step:Observe that the finiteness constraint is vacuous for a finite index set]
The index set $\{1, 2, \ldots, n\}$ is finite. In any element $\prod_{i=1}^n V_i$ of $\mathcal{B}_{\mathrm{box}}$, the set of indices $i$ for which $V_i \neq X_i$ is a subset of $\{1, \ldots, n\}$, hence finite (as every subset of a finite set is finite). Therefore the condition "$V_i = X_i$ for all but finitely many $i$" is automatically satisfied, and every element of $\mathcal{B}_{\mathrm{box}}$ belongs to $\mathcal{B}_{\mathrm{prod}}$.
[guided]
The product [topology](/page/Topology)'s defining constraint is that only finitely many factors $V_i$ may differ from the full space $X_i$. This constraint exists to ensure that the product topology satisfies the [Universal Property of the Product Topology](/theorems/962): a map into the product is [continuous](/page/Continuity) if and only if each coordinate function is continuous. For infinite products, the box topology is strictly finer and does not satisfy this universal property.
However, when the index set is $\{1, 2, \ldots, n\}$ — a finite set — any subset of indices where $V_i \neq X_i$ is automatically finite, regardless of what the $V_i$ are. The constraint "$V_i = X_i$ for all but finitely many $i$" is vacuous: there are only $n$ indices in total, so at most $n$ of them can have $V_i \neq X_i$, and $n$ is finite.
Consequently $\mathcal{B}_{\mathrm{box}} \subset \mathcal{B}_{\mathrm{prod}}$.
[/guided]
[/step]
[step:Conclude that the two bases, and hence the two topologies, coincide]
The inclusion $\mathcal{B}_{\mathrm{prod}} \subset \mathcal{B}_{\mathrm{box}}$ is immediate: every element of $\mathcal{B}_{\mathrm{prod}}$ is a product of [open sets](/page/Open%20Set), which is an element of $\mathcal{B}_{\mathrm{box}}$. Combined with the reverse inclusion $\mathcal{B}_{\mathrm{box}} \subset \mathcal{B}_{\mathrm{prod}}$ established in the previous step, we obtain $\mathcal{B}_{\mathrm{prod}} = \mathcal{B}_{\mathrm{box}}$. Since the two topologies are generated by the same basis, they are equal.
[/step]
