It is possible iff the set of the possible values of $x_{i}$ does not depend on the specific values of other $x_{j}$ in the sense that, given all values of $x_{j}$ for $j\neq i$ , the set of possible values of $x_{i}$ should be either the empty set or a fixed set $X_{i}$ (compare to Exercise 10 of §1). If this is true, then the set $X=\prod_{i\in\mathbb{Z}_{+}}X_{i}$ . So, we get the positive answer in (a), (b) and (c). Namely, we have $\mathbb{Z}^{\omega}$ , $(\overline{\mathbb{R}}_{+}+1)\times(\overline{\mathbb{R}}_{+}+2)\times\cdots$ where $\overline{\mathbb{R}}_{+}+k=\{x\in\mathbb{R}|x\ge k\}$ , and $\underbrace{\mathbb{R}\times\cdots\times\mathbb{R}}_{99}\times\mathbb{Z}\times\mathbb{Z}\times\cdots$ , respectively. In (d), when $x_{2}=c$ , $x_{3}$ can take values from $\{c\}$ only, so the set in (d) is not a cartesian product.