It is sufficient to show that $\mathbf{0}$ can be separated by a continuous function from a closed set $A$ that does not contain it. There is a neighborhood $W=\sqcap_j(-r_j,r_j)$ that does not intersect $A$ . Consider a linear transformation $h_j:x_j\to x_j/r_j$ . We have shown back in the exercises of Section 19 that the linear transformation is a homeomorphism in both the product and box topology on $\mathbb{R}^\omega$ . It seems a more general result is true: a product of continuous functions is continuous (so, that the product of homeomorphisms is a homeomorphism). If $\mathbf{y}=h(\mathbf{x})\in V$ , $V$ is open in the box topology, then there is a basis neighborhood $\mathbf{y}\in V'=\sqcap_j V'_j\subseteq V$ and $h(\mathbf{z})\in V'$ iff for all $j$ : $h_j(z_j)\in V'_j$ iff for all $j$ : $z_j\in h_j^{-1}(V'_j)$ . Therefore, for all $\mathbf{z}\in\sqcap_j h_j^{-1}(V'_j)$ which is open in the box topology if all $h_j$ are continuous, $h(\mathbf{z})\in V'\subseteq V$ . Anyway, it is true for the linear transformation, and if could separate $\mathbf{0}$ from $h(A)$ by a continuous function $f$ , then $f\circ h$ would separate $\mathbf{0}$ and $A$ . Note, that $h(A)$ is closed and does not have any points in $W'=\sqcap_j (-1,1)$ for if it had some than $A$ would have a point in $W$ . Now, since $\mathbb{R}^J_{uniform}$ is completely separable (as a metric space), there is a continuous function $f$ that separates $\mathbf{0}$ and $X-B(\mathbf{0},1)\subseteq W'$ (remember, that $W'$ is not open in the uniform metric: exercise 6 of Section 20). Now, $f$ is continous in the finer box topology as well, and separates $\mathbf{0}$ from $h(A)$ .