# $0$-dimensional and $G^{⋆⋆}$-regular

1 ) Why $X = \{ 0 \} \cup \{ 1 / n : n \in \mathbb{N} \}$ is $0$-dimensional ?

2 ) Let $X$ be a space and $G$ a topological group, Why If $X$ is $0$-dimensional in the sense of ind, then $X$ is $G^{**}$-regular ?

Note 1 : $X$ is $0$-dimensional if has a base of clopen sets.

Note 2 : $G^{⋆⋆}$-regular provided that, whenever $F$ is a closed subset of $X$, $x \in X \setminus F$ and $g \in G$, there exists $f \in Cp(X,G)$ such that $f(x) = g$ and $f(F) \subseteq \{e\}$.

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1) The sets of the form $\{\frac1n\}$ and $\{x\in X\mid x<\frac1n\}$ form a basis consisting of clopen sets.
2) $X\setminus F$ is an open neighbourhood of $x$, hence there is a smaller clopen neighbourhood $U$. Define $f(a)=g$ for $a\in U$, $f(a)=e$ otherwise.
 why $f$ is continuous? – Wreza Shafaghi Nov 3 '12 at 16:02 @WrezaShafaghi: $U$ is clopen, hence the inverse image of any (open) subset of $G$ is either $\emptyset$ or $U$ or $X\setminus U$ or $X$, thus always open. – Hagen von Eitzen Nov 3 '12 at 21:06 This is a complete solid answer Hagen +1. I agree with you in meta question. – Babak S. Nov 13 '12 at 16:05