$F$ defined as in the proof of the Theorem 34.1 maps $X$ to $\mathbb{R}^J$ . It is continuous as the range is in the product topology. It is injective because $X$ is a $T_1$ -space (for a pair of points there is a function in the family such that it maps the points to different values). We need to show that $F$ maps open sets in $X$ to open sets in the image. For $x\in U$ find an index $\alpha$ such that $f_\alpha(x)\neq 0=f_\alpha(X-U)$ . Then the set of points $y\in\mathbb{R}^J$ such that $y_\alpha\neq 0$ is an open neighborhood of $F(x)$ in $\mathbb{R}^J$ and its intersection $W$ with the image of $F$ is open in the image. Moreover, if $x'\notin U$ then $f_\alpha(x')=0$ and $F(x')\notin W$ , hence, $W\subseteq F(U)$ .