Proof of Lindelöf Theorem

Question

I have been surfing the net to read the proof of the Lindelöf Theorem:

Let $U\in \mathbb{R}^n$ be open and $U=\bigcup_{\lambda \in \Lambda} U_{\lambda}$where $\Lambda$ is an index set, $\{U_{\lambda}\}$ is a collection of open sets. Then, ther eis a countable subcollection $\{U_i\}$ of $\{U_{\lambda}\}$ so that $U=\bigcup_{i=1}^\infty U_i$.

I found out that most of the proof in the internet are from complex analysis. The level does not fit me as a beginner of real analysis, so I hereby ask for a more detailed and suitable proof. (Again, please don't lead me to some websites which talk about complex analysis!)

Thanks in advance.

Answer 1

Choose one open set $U_\lambda$ for every $n$-tuple with all rational coordinates.

Answer 2

Check that the collection $\mathcal{B}$ of open balls $B(x, r)$ of $\mathbb{R}^n$ where $x$ ranges over the points with all rational coordinates (of which there are only countably many) and $r>0$ ranges over the rational numbers, is a countable base of $\mathbb{R}^n$, which means that every open set $O$ of $\mathbb{R}^n$ can be written as a union of members from $\mathcal{B}$. Enumerate $\mathbb{B}$ as $\{B_n: n \in \mathbb{N}\}$.

For every $x \in U$, pick $\lambda(x) \in \Lambda$ such that $x \in U_{\lambda(x)}$. Then pick $n(x) \in \mathbb{N}$ such that $x \in B_{n(x)} \subseteq U_{\lambda(x)}$. For every $n$ in $I = \{n_x: x \in U \}$ (which is countable) pick one $\lambda(n)$ such that $\lambda(n) = \lambda(n_x)$ where $n_x = n$. Then $\{ U_{\lambda(n)} : n \in I \}$ is the required countable subfamily: pick $p \in U$, then we have $n(p)$ and $\lambda(p)$ such that $p \in B_{n(p)} \subseteq U_{\lambda(p)}$, so $n = n(p) \in I$, and we have $\lambda(n)$ to witness that $p \in U_{\lambda(n)}$, as required.

This could all be made explicit if we start with a concrete enumeration of the base, and take minimal index elements whenever we make choices above. The above is just a general proof of the fact that a second countable space is hereditarily Lindelöf...