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Is every regular separable space normal? Here are some (standard, I think) definitions:

  1. A space is called regular if given any nonempty closed set $F$ and any point $x$ that does not belong to $F$, there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $F$ that are disjoint.
  2. A space is called normal if, given any disjoint closed sets $E$ and $F$, there are open neighbourhoods $U$ of $E$ and $V$ of $F$ that are also disjoint.
  3. A space is called separable if it contains a countable, dense subset.
zuriel
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2 Answers2

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Another example is the square of the Sorgenfrey line $\mathbb{S}$ (the lower limit topology on $\mathbb{R}$, generated by the family of all half-open intervals of the form $[a,b)$ for $a < b$).

  • As $\mathbb{Q}$ is a dense subset of $\mathbb{S}$, it follows that $\mathbb{Q} \times \mathbb{Q}$ is dense in $\mathbb{S} \times \mathbb{S}$, and so it is separable.

  • $\mathbb{S}$ can be shown to be (perfectly) normal. As products of (completely) regular spaces are (completely) regular, it follows that $\mathbb{S} \times \mathbb{S}$ is (completely) regular.

  • To see that $\mathbb{S} \times \mathbb{S}$ is not normal is a bit more difficult. The "cheating" route is to apply Jones's Lemma:

    Jones' Lemma. Suppose $X$ is a normal space. Then given any closed discrete $F \subseteq X$ and any dense $D \subseteq X$ we have that $2^{|F|} \leq 2^{|D|}$.

    To apply this, as above $D = \mathbb{Q} \times \mathbb{Q}$ is a dense subset of $\mathbb{S} \times \mathbb{S}$, and it can be shown that the anti-diagonal $F = \{ \langle x , -x \rangle : x \in \mathbb{R} \}$ is a closed discrete subset. However $2^{|D|} = 2^{\aleph_0} < 2^{2^{\aleph_0}} = 2^{|F|}$. Therefore $\mathbb{S} \times \mathbb{S}$ cannot be normal.

    The less cheating method is to show that $F = \{ \langle x , -x \rangle : x \in \mathbb{Q} \}$ and $E = \{ \langle x , -x \rangle : x \in \mathbb{R} \setminus \mathbb{Q} \}$ are disjoint closed sets which cannot be separated by disjoint open sets. (An application of the Baire Category Theorem is useful here.)

Community
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user642796
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This is false. I consulted the excellent book Counterexamples in Topology and found example #103: An uncountable product (cardinality of product indices less than or equal to $2^{\aleph_0}$) of the natural numbers.

I recommend consulting your library for a copy of the book as the proofs of separability and normality are not one-liners.

kahen
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  • Thank you!! Then is the Tietze extension theorem true for every regular separable space? – zuriel Jun 11 '14 at 08:51
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    @zuriel No, when Tietze holds, the space is normal. But a regular separable space need not be normal. – Henno Brandsma Jun 11 '14 at 12:31
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    @kahen You can also try http://austinmohr.com/home/?page_id=146 for lookups of examples in that book. A handy resource! (it will give the Sorgenfrey square, the Niemytzki plane, and the rational sequence topology, as your example is not "unique" (there can be more such products). – Henno Brandsma Jun 11 '14 at 12:35