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Let $X$ be a normal space and let $U_{1}$, $U_{2}$ be open subsets of $X$ such that $X= U_{1} \cup U_{2}$. Show that there are open sets $V_{1}$ and $V_{2}$ such that $\overline{V_{1}} \subset U_{1}$, $\overline{V_{2}} \subset U_{2}$ and $X=V_{1} \cup V_{2}$.

What I've tried:

Note $X \setminus U_{2}$ is closed and $U_{1}$ is an open set containing it, hence by normality there is an open set $V_{1}$ such that $X \setminus U_{2} \subset V_{1} \subset \overline{V_{1}} \subset U_{1}$.

Similarly we can find an open set $V_{2}$ such that $X \setminus U_{1} \subset V_{2} \subset \overline{V_{2}} \subset U_{2}$.

Then from here I don't see how to conclude $X=V_{1} \cup V_{2}$. Perhaps this is a wrong approach. Can you please help?

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After you have defined $V_1$, notice that $X = V_1 \cup U_2$. Try defining $V_2$ with that observation in mind. –  yasmar Aug 10 '11 at 7:24

1 Answer 1

up vote 3 down vote accepted

To reduce notational clutter, let $F_1 = X \setminus U_1$ and $F_2 = X \setminus U_2$. You’ve chosen open sets $V_1,V_2$ such that $F_2 \subseteq V_1 \subseteq \operatorname{cl}V_1\subseteq U_1$ and $F_1 \subseteq V_2 \subseteq \operatorname{cl}V_2\subseteq U_2$, and you’d like to show that $X = V_1 \cup V_2$. Unfortunately, you can’t guarantee this. Take $X$ to be $[0,1]$ with the usual topology, $U_1 = [0,1)$, and $U_2 = (0,1]$. Then $F_1 = \{1\}$, $F_2 = \{0\}$, and $V_1$ and $V_2$ could turn out to be $[0,1/4)$ and $(3/4,1]$, for instance. This shows that your idea won’t work as it stands.

Notice that since $U_1 \cup U_2 = X$, $F_1 \cap F_2 = \varnothing$. That is, $F_1$ and $F_2$ are disjoint closed sets in the normal space $X$. Therefore there are open sets $V_1,V_2$ such that $F_1 \subseteq V_1$, $F_2 \subseteq V_2$, and $\operatorname{cl}V_1 \cap \operatorname{cl}V_2 = \varnothing$. Let $W_1 = X \setminus \operatorname{cl}V_1$ and $W_2 = X \setminus \operatorname{cl}V_2$; clearly these are open sets and $W_1 \cup W_2 = X$, so it only remains to show that $\operatorname{cl}W_1 \subseteq U_1$ and $\operatorname{cl}W_2 \subseteq U_2$, which isn’t too hard: clearly $X \setminus V_1$ is a closed set containing $W_1$, so $\operatorname{cl}W_1 \subseteq X \setminus V_1 \subseteq X \setminus F_1 = U_1$, and a similar computation shows that $\operatorname{cl}W_2 \subseteq U_2$.

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can we really guarantee that $cl(V_{1}) \cap cl(V_{2}) = \emptyset$? I thought that we can say that $cl(V_{1}) \cap V_{2} = \emptyset$ and $cl(V_{2}) \cap V_{1} = \emptyset$. Does this implies what you wrote? Can you please explain that line? –  user10 Aug 11 '11 at 2:12
    
@user10: By normality you can find disjoint open sets $G_1,G_2$ such that $F_1\subseteq G_1$ and $F_2\subseteq G_2$. Now apply normality again to find open sets $V_1,V_2$ such that $F_1\subseteq V_1\subseteq\operatorname{cl}V_1\subseteq G_1$ and $F_2\subseteq V_2\subseteq\operatorname{cl}V_2\subseteq G_2$. Then $\operatorname{cl}V_1\cap\operatorname{cl}V_2\subseteq G_1\cap G_2=\varnothing$. –  Brian M. Scott Aug 11 '11 at 2:44

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