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Definition: A Hausdorff space is normal (or: $T_4$) if each pair of disjoint closed sets have disjoint neighborhoods.

Then, we have

Exercise 5, pg. 158, Dugundji's Topology: Let $X$ be the upper of the Euclidean plane $E^2$, bounded by the $x$-axis. Use the Euclidean topology on $\{(x,y)\,|\, y>0\}$, but define neighborhoods of the points $(x,0)$ to be $\{(x,0)\}\cup [\text{open disc in $\{(x,y)\,|\,y>0\}$ tangent to the $x$-axis at $(x,0)$}]$. Prove that this space is not normal.

It is easy to see that $X$ is Hausdorff. So, what we need is to find a pair of closed sets that fail to satisfy the definition given above. Here, Alice Munro says that $A=\{(x,0)\,|\, x\in \Bbb Q\}$ and $B=\{(x,0)\,|\, x\in \Bbb R-\Bbb Q\}$ are such sets. I can see that they are closed, but how can I show that they do not admit disjoint neighborhoods? (intuitively true, but I'm having difficulty to write it down...)

Martin Sleziak
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Derso
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1 Answers1

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My preferred way (though the statement that these two sets are closed disjoint sets that cannot be separated is true) is to use a well-known cardinal number fact in normal spaces often called Jones' lemma:

Let $X$ be $T_4$. If $D$ is a dense subset of $X$ and $C$ is a closed and discrete (as a subspace) subset of $X$, then $2^{|C|} \le 2^{|D|}$.

I prove it in this answer; it uses Urysohn's extension theorem for normal spaces: we need a lot of different continuous real-valued functions on $X$ to separate all subsets of $C$, and $D$ bounds that number.

There I also show how it applies to the Moore plane (or Niemytzki plane): the $x$-axis is closed and discrete and $\mathbb{Q} \times \mathbb{Q}^+$ is countable and dense and $2^{|\mathbb{R}|} \not\le 2^{\aleph_0} = |\mathbb{R}|$.

Henno Brandsma
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