TITLE: Can you add a single point to any completely regular (not normal) space to get a normal space?
QUESTION [5 upvotes]: This is a follow up to my previous question (unfortunately closed as a duplicate).  There the problem was to turn the Moore plane into a normal space by adding a single point. Brian M. Scott gave an answer to this specific problem years ago.
This got me thinking about a more general question: Can you add one point to any completely regular not normal space $X$ to obtain a normal space $Y$ of which $X$ is a subspace?
Note that for me a completely regular spaces and normal spaces are Hausdorff. The restriction to completely regular spaces is natural since every subspace of a completely regular space is itself completely regular, and normal spaces are completely regular. Since $Y$ is supposed to be normal, the subspace $X$ must be completely regular.
I know that if $X$ is locally compact you can take $Y$ to be the one-point compactification. However if $X$ is not locally compact the one-point compactification fails to be Hausdorff.

REPLY [5 votes]: In general it is not possible to add a point to a Tychonoff space to get a normal space. Let $X= {\mathbb N}^\kappa$ where $\kappa$ is uncountable and where the topology is the product topology, that is, a base consists of products $\prod U_\lambda$ where finitely many sets $U_\lambda$ are single points and all others are all of $\mathbb N$. It is known that $X$ is not normal. Each basic open set is closed in $X$ and homeomorphic to $X$ (and therefore not normal).  Now let $Y = X \cup \{p\}$ be a Hausdorff space, where $X$ has the subspace topology. To show that $Y$ is not normal, it is enough to find a closed subset $K$ which is not normal.  Choose any $x \in X$ and let $U$ and $V$ be disjoint neighborhoods of $x$ and $p$, respectively.  Let $K = Y \setminus V$. Then $U,$ and, hence, $K$, contains a basic open set, which is closed and non-normal.  Therefore, $K$ is not normal.