Is there any general "mechanism" or rule-of-thumb to show that a subset of a metric space $(M,d)$ is dense. For example I am currently trying to digest the proof for:
$M$ satisfies second axiom of countability $\implies$ $M$ is separable
Let $(B_n)_{n\in \mathbb{N}}$ be a countable base for $M$ and choose $a_n ∈ B_n$ for all $n\in \mathbb{N}$. We only need to show that $A = \{a_n : n\in \mathbb{N}\}$ is dense in $M$. Given any non-empty open set $O$ and $x ∈ O$, there exists $n\in \mathbb{N}$ such that $x ∈ B_n ⊆ O$ by definition of base. Consequently, $a_n ∈ O$ and $A ∩ O \neq ∅$ follows
In what way is this proving density of $A$?
A set $A$ is dense in a metric space $M$ if every point $x\in M$ either belongs to $A$ or is a limit point of $A$. For $x$ to be a limit point of $A$ means that every neighborhood of $x$ contains a point from $A$ other than $x$ itself. So, by considering an arbitrary nonempty open set $O$ and showing that $A\cap O\neq \emptyset$ we are showing that all nieghborhoods of any point not in $A$ contain a point from $A$, i.e. all points are limit points and thus $\bar{A}=M$.