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1) This is example 5 (page 177 of Dugundji's book). Every subspace of the Sorgenfrey line is separable yet it is not second countable. I know how to prove it is not second countable. My question is: why is every subspace of the Sorgenfrey line separable? the Sorgenfrey line is separable but not metrizable, so I don't see why this follows immediately.

2) Is $\mathbb{R}^{\omega}$ a Lindelöf space? (here $\omega$ means the natural numbers). Can we simply say that $\mathbb{R}^{\omega}$ is metrizable being the countable product of a metrizable space, $\mathbb{R}$, and also separable since the countable product of separable spaces is separable so we have a separable and metrizable space, thus Lindelöf. What is another way of seeing this?

Now just for fun, what if we consider $\mathbb{R}^{\mathbb{R}}$. Is this Lindelöf?

Martin Sleziak
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user10
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  • The first question seems to be answered here: http://math.stackexchange.com/questions/15272/sorgenfrey-line-is-hereditarily-separable – Martin Sleziak Jul 08 '11 at 06:21
  • @Martin Sleziak: thanks, just found this too. Unfortunately I don't see how to characterize the points he says, I don't understand his argument. – user10 Jul 08 '11 at 06:23
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    @user10: To your fun question: It is not. Fact: If $J$ is uncountable, then $\mathbb R^J$ is regular, but not normal. Now Lindelöf + regular $\implies$ normal, hence $\mathbb R^J$ can't be Lindelöf for such $J$. – Sam Jul 08 '11 at 08:00

1 Answers1

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(1) Let $X \subseteq \mathbb R$. Let $\mathcal S$ be the subspace topology that $X$ inherits as a subspace of the Sorgenfrey line, and let $\mathcal E$ be the topology that $X$ inherits from $E^1$, the reals with the Euclidean topology. You know that $\langle X, \mathcal E \rangle$ is separable, so there is a countable $D_0 \subseteq X$ that is $\mathcal E$-dense in $X$. This $D_0$ is almost $\mathcal S$-dense in $X$ as well.

To see this, let $V = (x,y] \cap X$ (where $x,y \in \mathbb R$ with $x<y$ ) be a non-empty basic open set in $\langle X, \mathcal S \rangle$. Then either $(x,y) \cap X \ne \emptyset$, or $V = \{y\}$. $(x,y) \cap X \in \mathcal E$ - it's open in the Euclidean topology - so if it's non-empty, it contains a point of $D_0$, and therefore so does $V$. We only have a problem in the second case, when $V = \{y\}$. Clearly such a point $y$ is an isolated point in $X$, so it must belong to any dense subset of $X$. Let $D = D_0 \cup \{y \in X:y \text{ is an isolated point in }X\}$. It should be clear that $D$ is dense in $\langle X, \mathcal S \rangle$, so the only question is whether it's countable. This will be the case provided that $E = \{y \in X:y \text{ is an isolated point in }X\}$ is countable.

For each $y \in E$ there must be a real number $x_y<y$ such that $(x_y,y] \cap X = \{y\}$. Consider the intervals $(x_y,y]$ in $\mathbb R$; clearly they must be pairwise disjoint (otherwise one would contain at least two points of $X$), so each must contain a different rational number. There are only countably many rationals, so there are at most countably many intervals $(x_y,y]$ with $y \in E$, $E$ is therefore at most countable, and $D$ is a countable $\mathcal S$-dense subset of $X$.

(2) Yes, the argument that you suggest works. You can also use Theorems VIII.6.2(3) and VIII.6.3 to see that $\mathbb R^\omega$ is second countable and therefore Lindelöf. I'll have to think a bit about your last question.

Edit: Sam's already answered the question and given a reason. If you're interested in the details, which are a bit messy, look at Theorem 3 in this paper by A.H. Stone. If you take his uncountable set $\Lambda$ to be $\mathbb R$, the space $T$ in that theorem is $(\mathbb Z^+)^\mathbb R$, and it's easy to check that it's a closed subspace of $\mathbb R^\mathbb R$. Since normality is inherited by closed subspaces, non-normality of $T$ implies non-normality of $\mathbb R^\mathbb R$.

t.b.
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Brian M. Scott
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  • Last question: Sam gave an argument in a comment above. – t.b. Jul 08 '11 at 09:01
  • @Theo: I was hoping to find an argument for which I could give the details without getting too technical; the proof that $\omega^{\omega_1}$ is not normal is bit rugged for someone working through Dugundji. – Brian M. Scott Jul 08 '11 at 09:34
  • @Sam: What about the $x$ s.t. $x_j=1$, $x_\alpha=0$ for $\alpha\in J\setminus {j}$? Either I'm still asleep, or $x\in X\setminus \bigcup_{\alpha\in J\setminus {j}}B_\alpha$. – Brian M. Scott Jul 08 '11 at 20:58
  • @Brian: You are of course right. It seems to me, I was already asleep. – Sam Jul 09 '11 at 00:04
  • I initially was hoping I could exhibit more creativity than this, but one can just adapt Stone's idea: If $J$ is uncountable, we can consider the closed subspace $\mathbb Z ^J \subset \mathbb R^J$, and in this space once more the closed subspace $X = {x \text{ is injective on }\mathbb Z^J\setminus x^{-1}(0) }$. Now the collection $B_\alpha = {x(\alpha) = 0}$ is an open covering of $X$ and has no countable subcover - for we can find an injection $x: J_0 \to \mathbb Z$ into the integers minus the origin for any countable index set $J_0 \subset J$ (and map $J\setminus J_0$ into ${0}$). – Sam Jul 09 '11 at 00:34
  • @Sam: Yes, that's nicely straightforward (with the obvious correction that $x$ should be injective on $J \setminus x^{-1}(0)$). – Brian M. Scott Jul 09 '11 at 04:57