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In my Topology text it is written that E. Hewitt in 1946 made a regular space $X$ such that a function $f$ from $X$ into $\Bbb R$ is continuous if and only if it is constant but it did not prove this and moreover it did not give any reference about so that I thought to ask a question where I ask to explain Hewitt construction or where give some reference about it. Here a reference from my Topology text.

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So could someone help me, please?

Antonio Maria Di Mauro
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    Well if $X$ is a singleton then every function is continuous and every function is constant. I think you are missing some additional properties of $X$. – Clement Yung May 28 '22 at 14:18
  • @ClementYung I add an image from my Topology text. Anyway I think that the text assume implicitly that $X$ is not a singleton which moreover I do not know if it is regular - but surely if it is regular then it is regular vacuously. – Antonio Maria Di Mauro May 28 '22 at 14:24
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    A set with two points equipped with the indiscrete topology is a (non Hausdorff) example. So I wonder whether Hewitt's example is Hausdorff. – Ruy May 28 '22 at 14:37
  • @Ruy, Sorry but I am not sure to understand: could you explain better what you want mean, please? – Antonio Maria Di Mauro May 28 '22 at 14:40
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    Most likely, the author meant "a $T_3$ space" (which is regular+Hausdorff) instead of simply "regular." Check the definition of regularity in your text. As Wikipedia says "Although the definitions presented here for "regular" and "T3" are not uncommon, there is significant variation in the literature: some authors switch the definitions of "regular" and "T3" as they are used here, or use both terms interchangeably." – Moishe Kohan May 28 '22 at 14:58
  • @MoisheKohan Yes, you are right: so the text effectively for regular space meant a $T_1$ (not Hausdorff!!!) space such that if $F$ is closed and $x\notin F$ then there exists two disjoint open sets $A_F$ and $A_x$ such that $$F\subseteq A_F\quad\text{and}\quad x\in A_x$$ So with respec this definition the Ruy answer is not valid because an indiscrete space is not $T_1$, right? – Antonio Maria Di Mauro May 28 '22 at 15:32
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    Yes, Rui's example is invalid with your definition. You should edit the question to clarify the definition and unaccept the answer. – Moishe Kohan May 28 '22 at 15:34
  • @MoisheKohan Well if Ruy's example is valid for regular space but not for $T_3$ space I think I will put a specific question for $T_3$ space. – Antonio Maria Di Mauro May 28 '22 at 15:36
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    Your definition (regular in the usual sense+T1) implies Hausdorff. Ruy's example is valid under the usual regularity assumption but not the one used by your textbook. – Moishe Kohan May 28 '22 at 15:49
  • @MoisheKohan Oh, surely!!! – Antonio Maria Di Mauro May 28 '22 at 15:50

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Consider a set $X$ of cardinality 2 equipped withe the so called indiscrete topology, meaning that the open sets are only $X$ and $\emptyset$. Then $X$ is a regular space almost vacuously, that is, you can't find too many instances of a point outside a closed set.

Moreover it is clear that every continuous function from $X$ to $\mathbb R$ (or any other Hausdorff space) is constant.

Ruy
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