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It is usually stated that the (possibly uncountable) product of regular topological spaces is regular. However the only proof that I know of this fact seems to use the full axiom of choice :

See here (proof based on Mukres' Topology p.197).

Do we really need AC (does this imply AC) ? Would it hold in ZF, or in ZF + a weaker form of choice ?

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The standard proof that the product of regular spaces is regular does not use the Axiom of Choice.

Suppose $X_i$ ($i \in I$) are regular spaces and let $X = \prod_{i \in I} X_i$. Given a point $x \in X$ and a basic neighborhood $U = \prod_{i \in I} U_i$ of that point, we can always find a basic subneighborhood $V = \prod_{i \in I} V_i$ of $x$ such that $\overline{V} \subseteq U$. Indeed, when $U_i = X_i$ let $V_i = X_i$ too. For the finitely many $i \in I$ such that $U_i \neq X_i$, pick a neighborhood $V_i$ of $x_i$ such that $\overline{V}_i \subseteq U_i$. This only involves making finitely many choices, so the Axiom of Choice is not necessary.

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Francois, just tiny addition: the existence of the product space does of course require AC. I know it's silly to mention, but maybe that was the OP's real issue, 'uncountable product' an all... – Peter Krautzberger Feb 12 '11 at 16:20
Why would the existence of the product require choice? The non-emptiness of the product requires choice, but if it's empty, it's regular anyways. – Amit Kumar Gupta Feb 12 '11 at 16:30
Thanks for the addition, Amit; that's what I meant. – Peter Krautzberger Feb 12 '11 at 20:48

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