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?

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

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 $\operatorname{cl}W_2 \subseteq U_2$.

  • $\begingroup$ 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? $\endgroup$ – user10 Aug 11 '11 at 2:12
  • $\begingroup$ @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 $\operatorname{cl}V_1\cap\operatorname{cl}V_2\subseteq G_1\cap G_2=\varnothing$. $\endgroup$ – Brian M. Scott Aug 11 '11 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.