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How would I go about showing that a Tychonoff space $(T_{3.5})$ is finitely productive; or is there a counter-example disproving it? That is, if $A$ and $B$ are Tychonoff, then $A \times B$ is Tychonoff. In this context Tychonoff is defined as a space that is $T_1$ and where for any point $x$ and closed set $C$ such that $x \notin C$, there exists a continuous function with range $[0,1]$ such that $f(x) = 0$, and $f(C) = 1$.

I understand how to show that a regular $T_3$ space is finitely productive, however I have no idea how to show it for $T_{3.5}$.

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    $\begingroup$ This property is arbitrarily productive (i.e. any number of spaces). $\endgroup$ – Henno Brandsma Aug 8 '18 at 21:50
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You can find a proof in any book on general topology. In fact, the following is true: Let $X_\alpha$, $\alpha \in A$, be a family of Tychonoff spaces. Then $P = \Pi_{\alpha \in A} X_\alpha$ is a Tychonoff space.

We have to show that each $p \in P$ and each open neighborhood $U$ of $p$ in $P$ admits a continuous $f : P \to I$ such that $f(p) = 1$ and $f(x) = 0$ for $x \notin U$.

Choose finitely many $\alpha_i \in A$ and open neigborhoods $U_i$ of $p_{\alpha_i}$ in $X_{\alpha_i}$ such that $\bigcap_i \pi^{-1}_{\alpha_i}(U_i) \subset U$, where $\pi_{\alpha_i} : P \to X_{\alpha_i}$ denotes projection. There exist continuous $f_i : X_{\alpha_i} \to I$ such that $f_i(p_{\alpha_i}) = 1$ and $f_i(x_{\alpha_i}) = 0$ for $x_{\alpha_i} \notin U_{\alpha_i}$. Define $f : P \to I, f = \min_i\{ f_i \circ \pi_{\alpha_i} \}$. Then $f(p) = 1$ and $f(x) = 0$ for $x \notin \bigcap_i \pi^{-1}_{\alpha_i}(U_i)$, in particular for $x \notin U$.

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