Let \((X_1,d_1)\) and \((X_2,d_2)\) be metric spaces, and let \[ f:X_1\to X_2 \] be a continuous surjection such that \[ d_1(p,q)\le d_2(f(p),f(q)) \] for every \(p,q\in X_1\).
- If \(X_1\) is complete, must \(X_2\) be complete? Give a proof or counterexample.
- If \(X_2\) is complete, must \(X_1\) be complete? Give a proof or counterexample.
Solution
1. The map \(f\) is injective, hence bijective.
If \(f(p)=f(q)\), then the metric comparison gives \[ d_1(p,q) \le d_2(f(p),f(q)) =0. \] Thus \(p=q\). Since \(f\) is surjective by hypothesis, it is bijective.
2. If \(X_1\) is complete, then \(X_2\) is complete.
Let \((y_n)\) be a Cauchy sequence in \(X_2\). By step 1, define \[ x_n\coloneqq f^{-1}(y_n). \] Then \[ d_1(x_m,x_n) \le d_2(f(x_m),f(x_n)) = d_2(y_m,y_n), \] so \((x_n)\) is Cauchy in \(X_1\). If \(X_1\) is complete, there is \(x\in X_1\) with \[ x_n\longrightarrow x. \] Continuity of \(f\) gives \[ y_n=f(x_n)\longrightarrow f(x)\in X_2. \] Thus every Cauchy sequence in \(X_2\) converges.
3. Therefore the answer to part 1 is\[ \boxed{\text{yes}}. \]
This is exactly step 2.
4. For part 2, let\[ X_1=(0,1), \qquad X_2=\RR, \]with their usual metrics, and define\[ f(x)=\tan\bigl(\pi(x-1/2)\bigr). \]Then \(f:X_1\to X_2\) is a continuous surjection satisfying the required metric comparison.
The tangent function is continuous and strictly increasing from \(-\infty\) to \(\infty\) on \((-\pi/2,\pi/2)\), so the displayed \(f\) is a continuous bijection from \((0,1)\) onto \(\RR\).
Moreover, \[ f'(x) = \pi\sec^2\bigl(\pi(x-1/2)\bigr) \ge \pi >1. \] For \(x\ne y\), the mean value theorem gives some \(c\) between them such that \[ \abs{f(x)-f(y)} = f'(c)\abs{x-y} \ge \abs{x-y}. \] Thus \[ d_1(x,y)\le d_2(f(x),f(y)). \]
5. In the example from step 4, \(X_2\) is complete but \(X_1\) is not complete.
The real line with its usual metric is complete. The interval \((0,1)\) is not complete: for example, \[ x_n=\frac1n \] is Cauchy in \((0,1)\) but converges in \(\RR\) to \(0\notin(0,1)\).
6. Therefore the answer to part 2 is\[ \boxed{\text{no}}. \]
Steps 4 and 5 give a counterexample satisfying every hypothesis.
7. Q.E.D.
Step 3 proves part 1, and step 6 proves part 2.