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 $\map {B_{1/k} } {x_n} \cap Y$.
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:
\(\ds \map {d_Y} {y, y_{n, k} }\) | \(=\) | \(\ds \map {d_Y} {y, x_n} + \map {d_Y} {x_n, y_{n, k} }\) | Definition of Metric | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac 1 k\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
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 $Y$.
$\blacksquare$