Let $\mathit f:X_1 \to X_2$ be continuous and surjective. With certain property of $d$, if $(X_1, d_1)$ is complete, then is $(X_2,d_2)$ complete?

Question

Let $\mathit f:X_1 \to X_2$ be continuous and surjective, and $d_1(p,q)\le d_2 \bigl(\mathit f(p),\mathit f(q)\bigl)$, $\forall p,q\in X_1$.

1. If $(X_1, d_1)$ is complete, then is $(X_2,d_2)$ complete?

2. If $(X_2,d_2)$ is complete, then is $(X_1,d_1)$ complete?

I have proved the second question as below:

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Suppose that $(X_2,d_2)$ is complete. Then for a Cauchy sequence $\bigl(\mathit f(x_i)\bigl)_{i=1}^\infty$ must converge, i.e. $\forall \epsilon \gt0$, $ \exists N\in\mathbb N$ such that $\forall n\gt m\gt N$, $d_2\bigl(f(x_n),f(x_m)\bigl)\lt\epsilon$, and $\lim_{x\to\infty}f(x_i)=f(x)$.

Since $d_1(x_n,x_m)\le d_2 \bigl(\mathit f(x_n),\mathit f(x_m)\bigl)\lt\epsilon$, $(x)_{i=1}^\infty$ is also a Cauchy sequence.

Since $f$ is continuous, $\lim_{x\to\infty}f(x_i)=f(x)$ implies $\lim_{x\to\infty}x_i=x$.

Therefore Cauchy sequence $(x)_{i=1}^\infty$ converges, which implies that $(X_1,d_1)$ is complete.

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However firstly I do not think I showed that EVERY Cauchy sequence converges, and secondly for the question 1, I think it is not necessarily true but I can not find a proof or an counterexample either.

Could anyone help me improve the proof above and also share some hints or counterexamples for the question 1?

Answer 1

For question 2. you might want to consider $f: (-\frac{\pi}{2};\frac{\pi}{2}) \rightarrow \mathbb{R}, x \mapsto \tan(x)$ (both with the usual Eucledian metric $d$). As $\tan'(x) = 1 + \tan(x)^2 \geq 1$, we get by the mean value theorem $$ d(x,y) = \vert x -y \vert \leq \vert \tan(x) - \tan(y) \vert = d(f(x), f(y)). $$

Added: After seeing Surb's answer I'd like to offer another view point how to see what is going on for question 1. This is some abstract nonesense that someone might find interesting.

Note that $d_1(x,y) \leq d_2(f(x), f(y))$ implies that $f$ is injective and therefore bijective. Now we have $d_1(f^{-1}(x), f^{-1}(y)) \leq d_2(x, y)$, i.e. $f^{-1}$ is lipschitz continuous. As lipschitz functions map Cauchy sequences to Cauchy sequences, we can transport back the Cauchy sequences from $X_2$ to $X_1$ and use the completeness of $X_1$. The limit point of the pulled back Cauchy sequence will then be the limit point of the original sequence by the continuity of $f$.

Answer 2

For question 1.

Let $(y_n)\subset X_2$ be a Cauchy sequence. As $f$ is surjective, there exists $(x_n)\subset X_1$ such that $f(x_n)=y_n$ for all $n$. Now $d_1(u,v)\leq d_2(f(u),f(v))$ for all $u,v$ implies that $(x_n)$ is a Cauchy sequence. The completeness of $(X_1,d_1)$ implies the convergence of $(x_n)$ towards $x\in X_1$. Finally, the continuity of $f$ implies that $(y_n)=(f(x_n))$ converges towards $y=f(x)\in X_2$.

Answer 3

The answer to question 2 is  NO.

I will consider the standard metrics in $\ \mathbb R.$

 

Q2:   Let $\ X_2:=\mathbb N\ $ (positive integers). Let $\ X_1:=\{\frac 1n:n\in\mathbb N\}.\ $ Function

$$\forall_{n\in\mathbb N}\quad g\big(\frac 1n\big):=\ n $$

is continuous, $\ X_2\ $ is  complete but $\ X_1\ $ is  not.