Box Topology on Finite Product Space is Product Topology

Theorem
Let $n \in \N$.

For all $k \in \left\{ {1, \ldots, n}\right\}$, let $T_k = \left({X_k, \tau_k}\right)$ be topological spaces.

Let $\displaystyle X = \prod_{k \mathop = 1}^n X_k$ be the cartesian product of $X_1, \ldots, X_n$.

Then the box topology and the Tychonoff topology on $X$ are identical.

Proof
Denote the Tychonoff topology on $X$ as $\tau$, and the box topology on $X$ as $\tau'$.

Suppose that $U \in \tau'$.

Then there exists an index set $I$ such that:


 * $\displaystyle U = \bigcup_{ i \mathop \in I } \left({ U_{i,1} \times U_{i,2} \times \cdots \times U_{i,n} }\right)$

where $U_{i,k} \in \tau_k$ for all $i \in I, k \in \left\{ {1, \ldots, n}\right\}$.

When $\operatorname {pr}_k: X \to X_k$ denotes the $k$th projection on $X$, we have:


 * $\operatorname {pr}_k^{-1} \left({U_{i,k} }\right) = X_1 \times X_2 \times \cdots \times X_{k-1} \times U_{i,k} \times X_{k+1} \times \cdots \times X_n$

It follows that:


 * $\displaystyle U = \bigcup_{ i \mathop \in I } \left({U_{i,1} \times U_{i,2} \times \cdots \times U_{i,n} }\right) = \bigcup_{ i \mathop \in I } \bigcap_{ k \mathop = 1}^{n_i} \operatorname {pr}_k^{-1} \left({U_{i,k} }\right)$

By definition of Tychonoff topology, $U \in \tau$, so $\tau' \subseteq \tau$.

Suppose that $U \in \tau$.

Then there exists an index set $I$, so $U$ can be expressed as:

$U = \displaystyle \bigcup_{ i \mathop \in I } \bigcap_{ l \mathop = 1}^{m_i} \operatorname {pr}_{j_l}^{-1} \left({U_{i,l} }\right)$

where $m_i \in \N$, $j_1, j_2, \ldots, j_{m_i} \in \left\{ {1, \ldots, n}\right\}$, and $U_{i,l} \in \tau_{j_l}$ for all $i \in I, l \in \left\{ {1, \ldots, m_i}\right\}$.

Then:

As $\displaystyle \bigcap_{l: k \mathop = j_l} U_{i,l} \in \tau_k$, it follows that:


 * $\displaystyle \left({X_1 \times X_2 \times \cdots \times X_{k-1} \times \bigcap_{l: k \mathop = j_l} U_{i,l} \times X_{k+1} \times \cdots \times X_n}\right) \in \tau'$

By definition of topology, it follows that $U \in \tau'$, so $\tau \subseteq \tau'$.