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.

All in all:

By Lemma 1, $\left\{{R_x: x \in U}\right\}$ is countable as $\left\{{R_x: x \in U}\right\}$ is a set of open intervals with rational numbers as endpoints.

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


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


 * $\left\{{R_x: x \in U}\right\}$ equals the range of $\left\langle{R^i}\right\rangle_{i \mathop \in N}$.

From this follows:


 * $\left\{{R_x: x \in U}\right\} = \left\{{R^i: i \in N}\right\}$ as $\left\{{R^i: i \in N}\right\}$ equals the range of $\left\langle{R^i}\right\rangle_{i \mathop \in N}$.

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

Therefore, it is possible to require that $\left\langle{R^i}\right\rangle_{i \mathop \in N}$ lacks duplicates.

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

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

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


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

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 $x = \chi \left({i}\right)$

We find, for every $i$ in $N$ and $x = \chi \left({i}\right)$:


 * $O^i \in C$ 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 $\left\langle{R^i}\right\rangle_{i \mathop \in N}$ lacks duplicates.

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

Image of countable set under mapping is countable.

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

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

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

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

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

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

So, $\left\{{O^i: i \in N}\right\}$ covers $U$.

$S$ is covered by $C$.

This means that $S$ is a subset of $U$.

$U$ is a subset of the union of the elements of $\left\{{O^i: i \in N}\right\}$.

Therefore, $S$ is a subset of the union of the elements of $\left\{{O^i: i \in N}\right\}$.

Accordingly, $S$ is covered by $\left\{{O^i: i \in N}\right\}$ by the definition of cover.

$\left\{{O^i: i \in N}\right\}$ is a countable subset of $C$.

Therefore, $S$ is covered by a countable subset of $C$.

This finishes the proof of the theorem.

Lemma 1
Let $R$ be a real set of intervals with rational numbers as endpoints.

Let every interval in $R$ be of the same type of which there are four: $\left({\ldots \,.\,.\, \ldots}\right)$, $\left[{\ldots \,.\,.\, \ldots}\right]$, $\left[{\ldots \,.\,.\, \ldots}\right)$, and $\left({\ldots \,.\,.\, \ldots}\right]$.

Then $R$ is countable.

Proof
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.

The endpoint of an interval in $R$ is characterized by a rational number as every interval in $R$ is of the same type.

Therefore, the set consisting of the left hand endpoints of every interval in $R$ is countable.

Also, the set consisting of the right hand endpoints of every interval in $R$ 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 interval in $R$ is countable.

A subset of this cartesian product is in one-to-one correspondence with $R$.

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

$R$ is countable by Lemma 2 as $R$ is in one-to-one correspondence with a countable set.

Lemma 2
Let $S$ be countable set.

Let $T$ be a set.

Let $T$ be in one-to-one correspondence with $S$.

Then $T$ is countable.

Proof
$S$ is countable.

Therefore, $S$ is in one-to-one correspondence with a subset of the natural numbers by a definition of countable set.

$T$ is in one-to-one correspondence with $S$.

Therefore, $T$ is in one-to-one correspondence with a subset of the natural numbers by Composite of Bijections is Bijection.

Accordingly, $T$ is countable by a definition of countable.