Lindelöf's Lemma/Lemma 1

Theorem
Let $C$ be a set of open real sets.

Then there is a countable subset $D$ of $C$ such that:
 * $\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$

Lemma
Let $\ds U = \bigcup_{O \mathop \in C} O$.

Let $x$ be an arbitrary point in $U$.

Since $U$ is the union of the sets in $C$, the point $x$ belongs to a set in $C$.

Name such a set $O_x$.

Since $O_x$ is open, $O_x$ contains an open interval $I_x$ that contains $x$.

By Between two Real Numbers exists Rational Number, a rational number exists between the left hand endpoint of $I_x$ and $x$.

Also, a rational number exists between $x$ and the right hand endpoint of $I_x$.

Form an open interval $R_x$ that has two such rational numbers as endpoints.

All in all:

By Lemma, $\set {R_x: x \in U}$ is countable as $\set {R_x: x \in U}$ is a set of open intervals with rational numbers as endpoints.

By Countable Set equals Range of Sequence, the countability of $\set {R_x: x \in U}$ means that there exists a sequence $\sequence {R^i}_{i \mathop \in N}$ where:


 * $N$ is a subset of $\N$


 * $\set {R_x: x \in U}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.

From this follows:


 * $\set {R_x: x \in U} = \set {R^i: i \in N}$ as $\set {R^i: i \in N}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.

Two sequences that differ only by one of them having duplicates, have the same range.

Therefore, it is possible to require that $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.

Now, let $i$ be an arbitrary natural number in $N$.

Let $R^i$ be an element in $\set {R^i: i \in N}$.

There is an $x$ in $U$ such that:


 * $R_x = R^i$ as $\set {R_x: x \in U} = \set {R^i: i \in N}$

Also, we know that a set $O_x$ in $C$ exists such that:


 * $R_x \subset O_x$

The uniqueness of the elements of N makes it possible to define a mapping $\chi$ that sends $i$ to $x$.

This allows us to define, for every $i$ in $N$:


 * $O^i = O_x$ where $x = \map \chi i$

We find, for every $i$ in $N$ and $x = \map \chi i$:


 * $O^i \in C$ as $O_x \in C$


 * $R^i \subset O^i$ as $R_x \subset O_x$ and $R_x = R^i$ and $O_x = O^i$

Every $R^i$ where $i \in N$ is uniquely determined by $i$ as $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.

Therefore, it is possible to define a mapping from $\set {R^i: i \in N}$ to $\set {O^i: i \in N}$ that sends $R^i$ to $O^i$ for every $i$ in $N$.

Image of countable set under mapping is countable.

Therefore, $\set {O^i: i \in N}$ is countable as $\set {R^i: i \in N}$ is countable.

$\set {O^i: i \in N}$ is a subset of $C$ as $O^i \in C$ for every $i \in N$.

Therefore, $\set {O^i: i \in N}$ is a countable subset of $C$.

We find by focusing on $R_x$ for an $x$ in $U$:

So, $\set {R_x: x \in U}$ covers $U$.

We find by focusing on $O^i$ for an $i$ in $N$:

So, $D = \set {O^i: i \in N}$ satisfies the proposition of the theorem.