test
[proofplan]
The isometric embedding $\Phi: X \to Y$ identifies $X$ with the subspace $\Phi(X) \subset Y$. Since $Y$ is a separable [metric space](/page/Metric%20Space), [Subspaces of Separable Metrizable Spaces](/theorems/942) guarantees that $\Phi(X)$ is separable. We then transfer separability back to $X$: because $\Phi$ preserves distances, it is injective, so the inverse map $\Phi^{-1}: \Phi(X) \to X$ is well-defined and is itself an isometry. Applying $\Phi^{-1}$ to a [countable](/page/Countable%20Set) [dense subset](/page/Dense%20Subset) of $\Phi(X)$ produces a countable dense subset of $X$.
[/proofplan]
[step:Identify $\Phi(X)$ as a subspace of the separable metric space $Y$ and conclude that $\Phi(X)$ is separable]
The image $\Phi(X) \subset Y$ is a subspace of the [metric space](/page/Metric%20Space) $(Y, d_Y)$, equipped with the restricted metric $d_Y|_{\Phi(X) \times \Phi(X)}$. Since $(Y, d_Y)$ is separable and [metrizable](/page/Metrizable%20Space), the [Subspaces of Separable Metrizable Spaces](/theorems/942) theorem applies: $\Phi(X)$ is separable. Let $E = \{e_1, e_2, \ldots\} \subset \Phi(X)$ be a [countable](/page/Countable%20Set) [dense subset](/page/Dense%20Subset) of $\Phi(X)$.
[/step]
[step:Construct the inverse map $\Phi^{-1}: \Phi(X) \to X$ and pull back the countable dense subset]
Since $\Phi: X \to Y$ is an isometric embedding, it preserves distances: $d_Y(\Phi(x), \Phi(x')) = d_X(x, x')$ for all $x, x' \in X$. In particular, $\Phi$ is injective — if $\Phi(x) = \Phi(x')$, then $d_X(x, x') = d_Y(\Phi(x), \Phi(x')) = 0$, so $x = x'$. The map $\Phi$ therefore restricts to a bijection from $X$ onto $\Phi(X)$, and the inverse map
\begin{align*}
\Phi^{-1}: \Phi(X) &\to X \\
y &\mapsto \text{the unique } x \in X \text{ with } \Phi(x) = y
\end{align*}
is well-defined. Define the set
\begin{align*}
D := \Phi^{-1}(E) = \{\Phi^{-1}(e_1), \Phi^{-1}(e_2), \ldots\} \subset X.
\end{align*}
[guided]
The isometric embedding hypothesis does two things. First, it guarantees injectivity: an isometry maps distinct points to distinct points (since distinct points have positive distance, and distance is preserved). This ensures that $\Phi^{-1}$ is a well-defined function on the image $\Phi(X)$.
Second — and this is the step where the full strength of the isometry hypothesis is used — the inverse map $\Phi^{-1}$ is itself distance-preserving. For any $y, y' \in \Phi(X)$, write $y = \Phi(x)$ and $y' = \Phi(x')$ for the unique preimages $x, x' \in X$. Then:
\begin{align*}
d_X(\Phi^{-1}(y), \Phi^{-1}(y')) = d_X(x, x') = d_Y(\Phi(x), \Phi(x')) = d_Y(y, y').
\end{align*}
The first equality is the definition of $\Phi^{-1}$, the second is the isometry property of $\Phi$, and the third substitutes back $y = \Phi(x)$, $y' = \Phi(x')$. The inverse $\Phi^{-1}: (\Phi(X), d_Y) \to (X, d_X)$ is therefore an isometry, and in particular it is continuous. This [continuity](/page/Continuity) is what allows us to transfer density from $\Phi(X)$ back to $X$: the image of a [dense](/page/Dense%20Subset) set under a continuous surjection is dense in the codomain.
Why does this argument fail for a general (non-isometric) injection? A continuous injection $\Phi: X \to Y$ need not have a continuous inverse — the canonical example is the map $t \mapsto (\cos(2\pi t), \sin(2\pi t))$ from $[0, 1)$ to $S^1$, which is a continuous bijection onto its image but whose inverse is discontinuous at $(1, 0)$. Without continuity of $\Phi^{-1}$, pulling back a dense set from $\Phi(X)$ need not produce a dense set in $X$. The isometry condition eliminates this failure mode entirely: an isometry and its inverse are always Lipschitz continuous (with Lipschitz constant $1$), so density transfers in both directions.
[/guided]
[/step]
[step:Verify that $D$ is a countable dense subset of $X$]
**Countability.** The set $D = \Phi^{-1}(E)$ is [countable](/page/Countable%20Set) because $E$ is countable and $\Phi^{-1}$ is a function (it maps each element of $E$ to exactly one element of $X$).
**Density.** Let $x \in X$ and $\varepsilon > 0$. We must produce an element of $D$ within distance $\varepsilon$ of $x$. The point $\Phi(x) \in \Phi(X)$, and $E$ is [dense](/page/Dense%20Subset) in $\Phi(X)$, so there exists $e_k \in E$ with $d_Y(e_k, \Phi(x)) < \varepsilon$. Let $x_k := \Phi^{-1}(e_k) \in D$. Using the isometry property of $\Phi$:
\begin{align*}
d_X(x_k, x) = d_X(\Phi^{-1}(e_k), \Phi^{-1}(\Phi(x))) = d_Y(e_k, \Phi(x)) < \varepsilon.
\end{align*}
The first equality uses $\Phi^{-1}(\Phi(x)) = x$ (since $\Phi$ is injective), and the second equality uses the fact that $\Phi^{-1}$ is an isometry from $(\Phi(X), d_Y)$ to $(X, d_X)$, as established in the previous step.
Since $x \in X$ and $\varepsilon > 0$ were arbitrary, $D$ is dense in $X$. Therefore $(X, d_X)$ is separable.
[/step]
