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I've got the following question, and I'm having trouble with it. I was hoping that someone here could help me.

Show that by adding one point to the Moore plane $\mathbf{M}$ one can obtain a normal space of which $\mathbf{M}$ is a subspace.

My definition of the Moore plane is just like on Wikipedia. That is, as a set $\mathbf{M} = \{ (x,y) \in \mathbb{R}^2 : y \geq 0 \}$, the points $(x,y)$ for $y > 0$ have as basic open neighborhoods the usual open discs centered at $(x,y)$, and the points $(x,0)$ have as basic open neighborhoods open discs of radius $r>0$ centered at $(x,r)$ together with the point $(x,0)$.

I know that $\mathbf{M}$ is not normal (and for us normal implies Hausdorff), since, for example, the closed sets $A = \{ (x,0) : x \in \mathbb{Q} \}$ and $B = \{ (x,0) : x \in \mathbb{R} - \mathbb{Q} \}$ cannot be separated by disjoint open sets. I know that $A$ and $B$ cannot both be closed in this new space. But I'm having trouble figuring out how to add a point to $\mathbb{M}$ to get a normal space.

Any help would be appreciated!

Alex Ravsky
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Alice Munro
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  • If $M$ were locally compact, I'd have a candidate... (It is not, though) – Henno Brandsma Apr 29 '17 at 14:18
  • I guess that the space $(\Bbb M^,\tau)$, constructed as follows, is normal. Pick a point $p\not\in\Bbb M$ and define a topology $\tau$ on the set $\Bbb M^=\Bbb M\cup{p}$ consisting of all sets $U$ such that $U$ is open in $\Bbb M$ or $U\ni p$, $U\setminus{p}$ is open in $\Bbb M$ and a set $\Bbb R_0\setminus U$ is finite, where $\Bbb R_0=\Bbb R\times{0}$. – Alex Ravsky May 01 '17 at 06:06
  • Thanks, @AlexRavsky, that seems to make sense. I've been working on this and I think I can show that the space is normal with the assumption of regularity. But I can't seem to figure out how to show it is regular. Do you have any ideas? Or maybe a full answer? ;-) – Alice Munro May 04 '17 at 08:05

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