Separable Space satisfies Countable Chain Condition

Theorem
Let $T = \struct {S, \tau}$ be a separable topological space.

Then $T$ satisfies the countable chain condition.

Proof
In order to demonstrate that $T$ satisfies the countable chain condition, it is sufficient to demonstrate that every disjoint set of open sets of $T$ is countable.

Because $T$ is separable, there exists a subset $\set {y_n : n \in \N}$ of $S$ which is everywhere dense in $S$.

Now consider an indexed family $\family {U_j}_{j \mathop \in J}$ of non-empty open sets of $T$ such that:
 * $\forall i, j \in J, i \ne j: U_i \cap U_j = \O$

Using Equivalence of Definitions of Everywhere Dense this implies that for every $j \in J$ there has to exist $n_j \in \N$ such that $y_{n_j} \in U_j$.

This gives rise to a well-defined mapping $f: J \to \N$ via $\map f j := n_j$.

In particular $f$ is injective:

there were to exist $i, j \in J$, $i \ne j$ such that $n_i = n_j$.

Then:
 * $y_{n_i} \in U_i \cap U_j$

But the latter is the empty set by assumption.

From this contradiction it follows that $J$ is countable by definition.

This concludes the proof.