# Showing that a space is completely normal iff every subspace is normal

I'd like to know if I've been successful in my attempt below.

Prove that a space $X$ is completely normal if and only if every subspace of $X$ is normal.

$\Rightarrow$ Let $X$ be a completely normal space and let $Y$ be a subspace of $X$. Let $A,B \subseteq Y$ be disjoint closed subsets. Then clearly $\bar{A} \cap B = A \cap B = A \cap \bar{B} = \varnothing$, so by complete normality there are disjoint open sets $U,V \subseteq X$ such that $A \subseteq U$ and $B \subseteq V$. Taking $U \cap Y$ and $V \cap Y$, we have disjoint open sets in the subspace topology on $Y$ containing $A$ and $B$, respectively, in $Y$. It follows that $Y$ is normal.

$\Leftarrow$ Suppose every subspace of $X$ is normal and let $A,B \subseteq X$ be separated subsets so that $\bar{A} \cap B = A \cap \bar{B} = \varnothing$. Let $Y$ be a subspace containing $\bar{A}$ and $\bar{B}$. Since $Y$ is normal, there are disjoint open sets $U,V \subseteq Y$ such that $\bar{A} \subseteq U$ and $\bar{B} \subseteq V$. Then since $A \subseteq \bar{A}$ and $B \subseteq \bar{B}$, it follows that $X$ is completely normal.

Thanks.

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$\newcommand{\cl}{\operatorname{cl}}$Your $\Rightarrow$ argument is fine, save that it would be clearer if you said explicitly that $\overline A$ denotes closure in $X$. (This is why I prefer the notation $\cl A$, since it is so easily modified to show the space in which the closure is being taken: $\cl_XA$.)
You $\Leftarrow$ argument isn’t right: $X$ is a subspace containing $\cl A\cup\cl B$, so you’re not actually using the hereditary normality of $X$ at all. Let $Y=X\setminus(\cl_XA\cap\cl_XB)$; $\cl_YA$ and $\cl_YB$ are disjoint closed subsets of $Y$, and $Y$ is both normal and an open subset of $X$, so ... ?
Since $Y$ is normal, there are disjoint open sets containing $\cl_Y A$ and $\cl_Y B$. Does the fact that $Y$ is open give us that these open sets are subsets of $Y$, and so are not only open sets in the subspace topology on $Y$ but open sets in $X$ as well, giving us our result? – Alex Petzke Jan 31 '13 at 2:07
@Alex: Yes. Let $U$ be an open subset of $Y$. By definition $U=Y\cap V$ for some $V$ open in $X$. But $Y$ is itself open in $X$, so $U=Y\cap V$ is open in $X$. – Brian M. Scott Jan 31 '13 at 2:53