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Let $O$ be an open neighbourhood of $a\in \mathbb{R}$ and $V$ be an open neighbourhood of $a$ such that $\overline{V} \subset O$.

$V$ and $\mathbb{R} \setminus \overline{O}$ are disjoint (open) cozero-sets. I want to show that $V$ and $\mathbb{R}\setminus \overline{O}$ are completely separated. This question stems from my reading of Proposition 1.1 of

A.V. Arhangel'skii, Topological groups and $C$-embeddings, Top. Appl. vol.115, no.3 pp.265–289, doi:10.1016/S0166-8641(00)00073-0

I see several references to find a logical relation between open subset of space $X$ and completely separated sets, for example in Gillman and Jerison's Rings of continuous functions:

  1. Theorem. Two sets are completely separated if and only if they are contained in disjoint zero-sets.
  2. Urysohn's lemma. Any two disjoint closed sets in a normal space are completely separated.

But I can't find such a relation for this situation.

In answers to a previous question of mine, Brian M. Scott and Thomas E. explained that $A,B\subset X$ are completely separated iff their closures are.

How can I show that $V$ and $\mathbb{R}\backslash \overline{O}$ are completely separated?

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As the real line $\mathbb{R}$ is perfectly normal, it follows that the zero sets are exactly the closed sets. (In a normal space the zero sets are exactly the closed Gδ sets, and in a perfectly normal space all closed subsets are Gδ. See this other answer of mine.)

So to show that $V$ and $\mathbb{R} \setminus \overline{O}$ are completely separated, it suffices to find disjoint closed sets $F,E$ such that $V \subseteq F$ and $\mathbb{R} \setminus \overline{O} \subseteq E$.

Now just note that

  • $\overline{V}$ and $\mathbb{R} \setminus O$ are closed,
  • $V \subseteq \overline{V}$ and $\mathbb{R} \setminus \overline{O} \subseteq \mathbb{R} \setminus O$, and
  • as $\overline{V} \subseteq O$, it follows that $\overline{V} \cap ( \mathbb{R} \setminus O ) = \varnothing$.
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