Box Topology on Finite Product Space is Tychonoff 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 $\tau' \subseteq \tau$.
$\Box$
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:
\(\displaystyle U\) | \(=\) | \(\displaystyle \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \operatorname {pr}_{j_l}^{-1} \left({U_{i, l} }\right)\) | $\quad$ | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \bigcap_{l: k \mathop = j_l} \operatorname {pr}_k^{-1} \left({U_{i,l} }\right)\) | $\quad$ | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \left({X_1 \times X_2 \times \cdots \times X_{k-1} \times \bigcap_{l: k \mathop = j_l} U_{i,l} \times X_{k + 1} \times \cdots \times X_n}\right)\) | $\quad$ | $\quad$ |
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 $\tau \subseteq \tau'$.
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.): $\S 19$