Subspace of Separable Metric Space is Separable

Theorem
Let $\struct {X, d}$ be a separable metric space.

Let $\struct {Y, d_Y}$ be a metric subspace of $\struct {X, d}$.

Then $Y$ is separable.

Proof
If $Y = X$, we are done immediately.

Now consider the case $Y \ne X$.

Pick $x \in X \setminus Y$.

Let $\set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$.

Iterate over the pairs $\struct {n, k}$ where $n, k \in \N$.

Check whether:


 * $\map {B_{1/k} } {x_n} \cap Y \ne \O$

where $\map {B_{1/k} } {x_n}$ is the open ball of radius $r$, centered at $x_n$.

If the intersection is non-empty, let $y_{n, k}$ be any element of $\map {B_{1/k} } {x_n}$.

Otherwise, let $y_{n, k} = x$.

Now consider the set $\mathcal S = \set {y_{n, k} : y_{n, k} \ne x}$.

Note that it is not yet known that $\mathcal S \ne \O$.

We show that $\mathcal S$. is everywhere dense.

Let $y \in Y$ and let $\epsilon > 0$ be arbitrary.

Let $k$ be such that $1/k < \epsilon/2$.

Since $\set {x_n : n \in \N}$ is everywhere dense, there exists $x_n \in \set {x_n : n \in \N}$ such that:


 * $\ds \map d {y, x_n} < 1/k$

Now, we have that:


 * $\map {B_{1/k} } {x_n} \cap Y \supseteq \set {y_{n, k} } \ne \O$

We are now assured that $\mathcal S \ne \O$.

So, we have:

So, from Condition for Point being in Closure, we have:


 * $y \in \mathcal S^-$

But $y \in Y$ was arbitrary, so we have:


 * $Y \subseteq \mathcal S^-$

So:


 * $Y = \mathcal S^-$

That is:


 * $\mathcal S$ is everywhere dense in $Y$.

We now show that $\mathcal S$ is countable.

From Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering and the Well-Ordering Principle, there exists a well-order $\sqsubseteq$ on $\N \times \N$.

For each $y \in \mathcal S$, define a map $f : \mathcal S \to \N \times \N$ by picking $\map f y$ to be the $\sqsubseteq$-least element of $\set {y_{n, k} \in \mathcal S : y_{n, k} = y}$.

Then $f$ is an injection.

From Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable.

So, from Domain of Injection to Countable Set is Countable, we have that:


 * $\mathcal S$ is countable.

So $\mathcal S$ is a countable everywhere dense subset of $Y$.