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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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up vote 5 down vote accepted

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.

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This is a great example to keep in mind to avoid this and other common errors. – Nate Eldredge Feb 17 '13 at 17:31

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