Prove that if two dense subsets of $X_1$ and $X_2$ are isometric, then $X_1$ and $X_2$ are isometric.

Question

Let $(X_1, d_1)$ and $(X_2, d_2)$ be complete metric spaces. Suppose that $Y_1 \subseteq X_1$ is dense in $X_1, Y_2 ⊆ X_2$ is dense in $X_2$, and there exists an isometry $f : Y_1 → Y_2$, where $Y_1, Y_2$ are endowed with the corresponding subspace metrics. Prove that there exists an isometry $F : X_1 → X_2$.

Answer 1

Hint: Try extending $f$ by taking limits. $f$ sends Cauchy sequences to Cauchy sequences (since it's an isometry), which converge (by completeness).

If $x\in X_1\setminus Y_1$, take a sequence $(y_n)$ in $Y_1$ such that $y_n\to x$. Define $\hat f: X_1\to X_2$ by $\hat f(x)=\begin{cases} f(x), x\in Y_1\\\lim_{n\to\infty} f(y_n), x\not\in Y_1\end{cases}$.