6

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.

Alice Munro
  • 113
  • 4
  • 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. – Henno Brandsma May 05 '17 at 18:42
  • @HennoBrandsma Wouldn't that mean that all singletons from $X$ are no longer closed in $Y$? – Alice Munro May 05 '17 at 21:22
  • No, only add points to all counterexamples to normality, and singletons are never part of those, by complete regularity of $X$ – Henno Brandsma May 05 '17 at 21:55
  • 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. – Henno Brandsma May 05 '17 at 21:57

1 Answers1

6

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.

Anonymous
  • 196
  • 2