# If there exist disjoint open sets containing separated sets, then the space is completely normal.

I am asked to prove subject along with its converse (which I can do). I would appreciate any assistance based on my progress below in proving the separation conditions hold in the situation I have constructed. Thanks!

DEFINITIONS

$\textbf{completely normal}$: every subspace is normal

$\textbf{separated}$: $A$ and $B$ are separated if and only if $\overline{A}\cap B=\varnothing=A\cap\overline{B}$

PROVE: If for every pair of separated sets $A$ and $B$ in $X$ there exist disjoint open sets containing them, then $X$ is completely normal.

IDEAS:

Let $Y$ be a subspace of $X$ and let $A$ and $B$ be disjoint closed sets in $Y$. It must be shown there exist disjoint open sets $U$ and $V$ in $Y$ containing $A$ and $B$ (respectively). First show $A$ an $B$ are separated.

$\textit{This is where I need help...}$

idea 1: Since $A$ and $B$ are closed in $Y$, there exist closed sets $A_X$ and $B_X$ in $X$ such that $A_X\cap Y=A$ and $B_X\cap Y=B$. Since $A_X$ and $B_Y$ are closed in $X$, then $\overline{A_X}=A_X$ and $\overline{B_X}=B_X$.

idea 2: The sets $A$ and $B$ may not be closed in $X$, so let $\overline{A}$ and $\overline{B}$ be their closures in $X$. Then $\overline{A}\cap Y=A$ and $\overline{B}\cap Y=B$.

$\textit{This is where I need to prove$A$and$B$are separated...}$

Since $A$ and $B$ are separated, there exist disjoint open sets $U'$ and $V'$ in $X$ containing $A$ and $B$ respectively by hypothesis. Then $U=U'\cap Y$ and $V=V'\cap Y$ are disjoint open sets in $Y$ containing $A$ and $B$ respectively so $Y$ is normal and $X$ is completely normal since $Y$ was an arbitrary subset of $X$.

• Does this make sense? Let $\overline{A}$ and $\overline{B}$ be the closures of $A$ and $B$ in $X$. Then $A=\overline{A}\cap Y$ and $B=\overline{B}\cap Y$. Since $A\cap B=\varnothing$, then $A\cap B=(\overline{A}\cap Y)\cap B=\overline{A}\cap(Y\cap B)=\overline{A}\cap B=\varnothing$ where $Y\cap B=B$ because $B$ is completely contained in $Y$. Use the same process for the other desired condition. May 13, 2015 at 18:07
• Yes, what you have in your comment is exactly what you need in order to complete the proof. May 13, 2015 at 22:12
• Excellent! Thank you so very much! May 14, 2015 at 21:21
• You’re welcome! May 14, 2015 at 21:27

Let $A,B$ be closed in $Y$ then $A=U\cap Y ,B=V\cap Y$ $U,V$ are closed in $X$
Also $\overline U_X\cap V=U\cap V=\emptyset;\overline V_X\cap U=U\cap V=\emptyset$
Thus $U,V$ are weakly separated in $X$ Thus $\exists$ open sets $P,Q\in X$ such that $U\subset P;V\subset Q;P\cap Q=\emptyset$
Thus $U\cap Y\subset P\cap Y;V\cap Y\subset Q\cap Y$ where $P\cap Y,Q\cap Y$ open in $Y$