Box Topology contains Product Topology

Theorem

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:

$\ds X := \prod_{i \mathop \in I} X_i$

Let $\tau$ be the product topology on $X$.

Let $\tau'$ be the box topology on $X$.

Then:

$\tau \subseteq \tau'$

Proof

From Natural Basis of Product Topology and Basis for Box Topology, it follows immediately that the natural basis $\BB$ for the product topology is contained in the basis $\BB'$ for the box topology.

From Corollary of Basis Condition for Coarser Topology, it follows that $\tau \subseteq \tau'$.

$\blacksquare$