Lindelöf's Lemma

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

Let $S$ be a set that is covered by $C$.

Then there is a countable subset of $C$ that covers $S$.

Proof
Let $U = \displaystyle \bigcup_{O \mathop \in C} O$.

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

Since $U$ is the union of the sets in $C$, $x$ 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, two rational numbers exist between $x$ and each endpoint of $I_x$.

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

By Rational Numbers are Countably Infinite, we have that the rationals are countable.

By Finite Union of Countable Sets is Countable, the pair of two countable sets is countable.

$R_x$ is defined from its two endpoints, which are rational numbers.

Therefore, the set $\left\{ {R_y: y \in U} \right\}$ is countable.

This allows us to define a different index $i$ for $R_x$ like this:


 * $R^i = R_x$

where


 * $i \in N$


 * $N \subseteq \N$

Note that:
 * $\left\{ {R^j: j \in N} \right\} = \left\{ {R_y: y \in U} \right\}$

Starting with $R^i$, we know that a $y$ in $U$ exists so that $R_y = R^i$.

Also, $R_y \subset I_y$ and $I_y \subset O_y$.

Define $O^i = O_y$.

We observe that $R^i \subset O^i$.

We define a mapping from $\left\{ {R^j: j \in N} \right\}$ to $\left\{ {O^j: j \in N} \right\}$ that sends $R^i$ to $O^i$.

By Image of Countable Set under Mapping is Countable, $\left\{ {O^j: j \in N} \right\}$ is countable since $\left\{ {R^j: j \in N} \right\}$ is countable.

We observe that $\left\{ {O^j: j \in N} \right\}$ is a countable subset of $C$.

Putting all this together:

Since $S$ is covered by $C$, we have $S \subset U$.

Also, $S \subset \displaystyle \bigcup_{i \mathop \in N} \left\{ {O^i} \right\}$ since:
 * $U = \displaystyle \bigcup_{i \mathop \in N} \left\{ {O^i} \right\}$

In other words, the collection $\left\{ {O^i: i \in N} \right\}$ covers $S$.