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.

  • $\begingroup$ Maybe add a point $p$ such that for all closed sets $A$ in $X$, $p$ is in its closure in $Y =X \cup \{p\}$. This would "kill" all non-normality showing pairs (which then cannot be closed and disjoint any more). ? Just an idea that might or might not work. $\endgroup$ – Henno Brandsma May 5 '17 at 18:42
  • $\begingroup$ @HennoBrandsma Wouldn't that mean that all singletons from $X$ are no longer closed in $Y$? $\endgroup$ – Alice Munro May 5 '17 at 21:22
  • $\begingroup$ No, only add points to all counterexamples to normality, and singletons are never part of those, by complete regularity of $X$ $\endgroup$ – Henno Brandsma May 5 '17 at 21:55
  • $\begingroup$ Brian M Scott did a similar thing: all counterexamples to normality came from the $x$-axis, so he made the new point a 1-point that would be a limit point to all those sets. You had a similar idea expressed by Ravski, as a comment to your new duplicate question, IIRC. $\endgroup$ – Henno Brandsma May 5 '17 at 21:57

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.

  • $\begingroup$ Nicely done. I would add that the basic open set that is a subset of $U$ is closed in $Y$ because it is closed in $X$ and is disjoint from the nbhd $V$ of $p.$ $\endgroup$ – DanielWainfleet May 6 '17 at 18:37
  • $\begingroup$ Thanks a lot, this is great! $\endgroup$ – Alice Munro May 26 '17 at 13:32

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