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?
Best Answer
(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 $\mathbb R^\mathbb R$.