Exercise 32.4

Question

Show that every regular Lindelöf space is normal.

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $A,B$ be disjoint closed subsets of $X$ regular and Lindelöf. For all $x \in A$, there exists a neighborhood $\mathcal{U} 
i x$ disjoint from $B$. By regularity, there exists a neighborhood $U \subseteq \overline{U} \subseteq \mathcal{U}$ containing $x$; since these $U$ cover $A$, and $A$ is Lindelöf by Exercise \S30.9, there exists a countable subcover $\{U_i\}$ such that $\overline{U_i} \cap B = \emptyset$ for all $i$. Similarly, we can construct a countable subcover $\{V_i\}$ of $B$ such that $\overline{V_i} \cap A = \emptyset$ for all $i$. By the exact same argument as in the proof of Theorem $32.1$, then, the sets
  \begin{equation*}
    U' = \bigcup_{n \in \mathbb{Z}_+} \bigg( U_n \setminus \bigcup_{i=1}^n \overline{V_i} \bigg), ~ V' = \bigcup_{n \in \mathbb{Z}_+} \bigg( V_n \setminus \bigcup_{i=1}^n \overline{U_i} \bigg)
  \end{equation*}
  are open and $U' \supseteq A, V' \supseteq B, U' \cap V' = \emptyset$, and so $X$ is normal.
\end{proof}

Exercise 32.4

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Exercise 32.4

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