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 $C$, 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.

By Rational Numbers are Countably Infinite, the rationals are countable.

By Subset of Countably Infinite Set is Countable, a subset of the rationals is countable.

Therefore, the set consisting of the left hand endpoints of every $R_x$ for every $x$ in $U$ is countable.

Also, the set consisting of the right hand endpoints of every $R_x$ for every $x$ in $U$ is countable.

The cartesian product of countable sets is countable.

Therefore, the cartesian product of the sets consisting of the respectively left hand and right hand endpoints of every $R_x$ for every $x$ in $U$ is countable.

A subset of this cartesian product is in one-to-one correspondence with the set of intervals $R_x$ for every $x$ in $U$.

This subset is countable by Subset of Countably Infinite Set is Countable.

That this subset is countable means that it is in one-to-one correspondence with a subset of the natural numbers by the definitions of countable set and bijection.

This subset is also in one-to-one correspondence with the set of intervals $R_x$ for every $x$ in $U$.

Therefore, the set of intervals $R_x$ for every $x$ in $U$ is in one-to-one correspondence with a subset of the natural numbers.

Therefore, the set of intervals $R_x$ for every $x$ in $U$ is countable by one of the definitions of countable.

The countability of the set of intervals $R_x$ for every $x$ in $U$ 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 $S$.