# 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?

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:

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.

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?

• $f$ being continuous and $\lim f(x_i) = f(x)$ does not imply $\lim x_i = x$. Take for example any constant function $f$ and a sequence $(x_i)$, s.t. $\lim x_i \neq x$. Commented Sep 29, 2019 at 20:24
• @KeeperOfSecrets Thanks a lot for helping me find the flaw. Commented Sep 29, 2019 at 22:44

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 $$y=f(x)\in X_2$$.

• Nice! Now I understand it much better +1 Commented Sep 29, 2019 at 20:46
• In my opinion, you still did not show that EVERY Cauchy sequence converges, did you? Why can you claim that the $(x_n)$ you obtained were all Cauchy sequences in $X_1$? Commented Sep 29, 2019 at 22:41
• Because $$d_1(x_n, x_m) \leq d_2 (f(x_n), f(x_m)) = d_2(y_n, y_m).$$ As $(y_n)$ is Cauchy, so is $(x_n)$. Commented Sep 30, 2019 at 7:36
• @H.Zhu As Severin said, $(x_n)$ is Cauchy, because $(y_n)$ is Cauchy and $d(x_n,x_m)\leq d(y_n,y_m)$ by the inequality involving $f$. By the way, I'd like to suggest you to fill in the details in my answer to better understand what happens.
– Surb
Commented Sep 30, 2019 at 8:38
• @Surb I meant that why can you be sure that there are no other Cauchy sequences in $X_1$ besides the $(x_n)$ you obtained that way? Our purpose should be to porve that EVERY Cauchy sequence converges right? But how can you guarantee that you have obtained all of them in that way? Commented Sep 30, 2019 at 9:00

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 $$f$$.

• I did not see any flaws in your added part... But there must be some since you have already showed a counterexample. Commented Sep 29, 2019 at 22:44
• Not really. My counterexample is for question 2. The proof is for question 1. You pick some Cauchy sequence $(y_n)$ in $X_2$. Then (as $f^{-1}$ is lipschitz) also $(f^{-1}(y_n))$ is Cauchy in $X_1$. However, $X_1$ is complete, thus there exists $x\in X_1$ such that $f^{-1}(y_n) \rightarrow x$. Then by the continuity of $f$ we have $$y_n =f(f^{-1}(y_n)) \rightarrow f(x)$$ Thus, every Cauchy sequence in $X_2$ converges. Commented Sep 30, 2019 at 7:30
• I did not say anything new, I just reformulated Surb's nice answer in the way I think about this problem. Commented Sep 30, 2019 at 7:32

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.

• It is a good example. Thanks a lot! Commented Sep 29, 2019 at 22:42