# Tychonoff Topology and Closure Operator.

After the 2.3.6 corollary in Engelking's General Topology, there is an observation alerting the reader that the Tychonoff Topology cannot be generated by the closure operator defined by $$\overline{\prod_{s\in S} A_{s}} = \prod_{s\in S}\overline{A}_{s},$$

for not all subset $B$ of the space $\prod_{s\in S} X_s$ can be represented in the form $\prod_{s\in S} B_s$, such that, for all $s\in S$, $B_s\subseteq X_s$.

But, if ${p}_{s}: \prod_{s\in S} X_s \to X_s$ are the projections and given $B\subseteq \prod_{s\in S} X_s$, it is not true that $B = \prod_{s\in S} {p}_{s}[B]$?

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No, consider the diagonal $D = \{(x,x): x \in \mathbb{R} \} \subset \mathbb{R}^2$. Here $\pi_1[D] = \pi_2[D] = \mathbb{R}$ and $D \neq \mathbb{R}^2$. $D$ is an example of a (closed) set that we cannot write as the product of two closed sets in the factors.