$\mathbb{R^n}$ Second Countable

Lemma: Let U be an open subset of $$\mathbb{R}^n$$ and $$x \in U$$ then there exists a ball with rational center $$(\mathbb{Q}^n$$ and rational radius containing x, that is contained in U.

What I want to show: Every open subset of $$\mathbb{R}^n$$ is the union of some some collection of elements in the following set:

$$\{$$ Open balls with rational center and rational radius $$\}$$

Let U be an arbitrary open set in $$\mathbb{R}^n$$. By the above lemma, for each $$x \in U$$ you can find a rational center $$x'$$ and rational radius $$r'>0$$ so that $$x \in B(x',r')$$ $$\subset U$$. Hence the following collection:

$$B'=$$ $$\{$$ B(a,r) : a $$\in \mathbb{Q}^n$$ , $$r>0$$ , B(a,r) $$\subset U$$ $$\}$$ gives us

$$U=$$ $$\bigcup_{B \in B'}$$ $$B$$

is the proof correct?

As this is part of the proof of showing that $$\mathbb{R}^n$$ is second countable, it follows that since every metric space is Hausdorff, $$\mathbb{R}^n$$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $$\mathbb{R}^n$$ is a n- dimensional topological manifold, am I correct?

• I assume that you require manifolds to be second countable? Then yes and yes. – freakish Aug 5 at 18:39
• @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ – topologicalmagician Aug 5 at 18:40
• @topologicalmagician You prove correctly that $\mathbb{R}^n$ is second countable. Also, since $\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. – bing-nagata-smirnov Aug 5 at 19:03

Yes, your proofs that $$\mathbb R^n$$ is cecond countable and that $$\mathbb R^n$$ is an $$n$$- dimensional topological manifold are correct.