The problem in Kreyzig states:

If $(X,d)$ is complete, show that $(X,\tilde{d})$, where $\tilde{d}=d/(1+d)$, is complete.

My attempt to solve it:

1) $(X,d)$ is complete, so any Cauchy sequence $\{x\}_{n=0}^\infty$ converges and $\forall \epsilon > 0,\; \exists N\in \mathbb{N}$ such that $\forall n,\forall m > N,\; m,n\in \mathbb{N}$ we have $d(x_n,x_m)<\epsilon$.

2) By the property of a metric $d(x,y)\geq0, \; \forall x,y \in X$.

3) $\tilde{d}(x_n,x_m) = d(x_n,x_m)/(1+d(x_n,x_m)) \leq d(x_n,x_m)<\epsilon$ so the arbitrary Cauchy sequence in $X$ converges with the metric $\tilde{d}$.

Is the proof correct or there are some issues with it? Is there any other way to prove it?

Thank you for your help in advance.

  • $\begingroup$ What is your $y_n$ in 3? You have one Cauchy sequence $\{x_n\}$ in $(X, \tilde d)$, you have to show that it converges. $\endgroup$ – user99914 Sep 22 '17 at 18:18
  • $\begingroup$ How does your 3) prove that the sequence is convergent - under either metric? $\endgroup$ – mathguy Sep 22 '17 at 18:19
  • $\begingroup$ @John Ma Corrected it, thank you! $\endgroup$ – Konstantin Sep 22 '17 at 18:20
  • $\begingroup$ If you use the definitions of "Cauchy sequence" and "convergent sequence" correctly, you should be able to show that every sequence that is Cauchy under $\tilde d$ is also Cauchy under $d$, and every sequence that is convergent under $d$ is also convergent under $\tilde d$. The conclusion will follow from these two observations, plus the original space being complete. $\endgroup$ – mathguy Sep 22 '17 at 18:21
  • $\begingroup$ Your logic is backwards. $\endgroup$ – DanielWainfleet Sep 22 '17 at 21:49

You want to show that: if $\{x_n\}$ is a Cauchy sequence in $(X, \tilde d)$, then $x_n \to x$ (in the metric $\tilde d$) for some $x\in X$.

So choose such a sequence $\{x_n\}$. Then for all $\epsilon >0$ and $\epsilon<1/2$, there is $N$ so that

$$ \tilde d (x_n, x_m) <\epsilon$$

for all $n, m\ge N$. Then

$$ \frac{d(x_n, x_m)}{1 + d(x_n, x_m)} <\epsilon.$$

Moving terms around, we have

$$ d(x_n, x_m) < \frac{\epsilon}{1-\epsilon}< 2\epsilon.$$

(we used $\epsilon<1/2$ in the last inequality). Thus $\{x_n\}$ is also a Cauchy sequence in $(X, d)$ and so $x_n\to x $ in the metric $d$. That is, for all $\epsilon >0$, there is $N$ so that

$$ d(x_n, x)<\epsilon$$

for all $n\ge N$. But since

$$ \tilde d(x_n , x) \le d(x_n, x) <\epsilon,$$

this shows that $\{x_n\}$ also converge to $x$ in the metric $\tilde d$.

  • $\begingroup$ Thank you very much for the outstanding reply! Your point that we have to start with a Cauchy sequence in $(X,\tilde{d})$ is exactly right! $\endgroup$ – Konstantin Sep 22 '17 at 18:48

From hypothesis the space $(X,d)$ is complete with respect to $d$,but you have to prove that $(X,\bar{d})$ is complete with respect to $\bar{d}$

Lets take a Cauchy sequence $x_n \in (X,\bar{d})$ and prove that $x_n$ is Cauchy with respect to $d$.

We have that exists $n_0 \in \Bbb{N}$ such that $\frac{d(x_n,x_m)}{1+d(x_n,x_m)}<\epsilon,\forall n,m \geq n_0$

We have that $d(x_n,x_m) \leq (1 +d(x_n,x_m))\epsilon \Rightarrow d(x_n,x_m) \leq \frac{\epsilon}{1-\epsilon},\forall n,m \geq n_0$

Thus $\limsup_{\epsilon \to 0}d(x_n,x_m) \leq 0 $

Using this you can prove that $\lim_{m,n} d(x_n,x_m)=0$

Now you have to prove that $x_n$ converges with respect to the new metric.

From the fact than $(X,d)$ is complete exists $x \in X$ such that $d(x_n,x) \to 0$.

But $0 \leq \bar{d}(x_n,x) \leq d(x_n,x) \to 0$

Thus $x_n \to^{\bar{d}} x$ , so $(X,\bar{d})$ is complete.


Your third line is a bit off. You shouldn't have $\tilde{d}(x_n,y_m)$. You want to look at elements of the same sequence: so, $\tilde{d}(x_n,x_m)$. The other thing you have that is suspect is $\tilde{d}(x_n,y_m)$ (sic) $\le d(x_n,y_n)$. The proper conclusion would be this: $\tilde{d}(x_n,x_m) \le d(x_n, x_m)$.

Now since $(X,d)$ is complete, we have $x_n \to x$ for some $x\in X$ (w.r.t the $d$ metric). But by what you've shown, this implies $\tilde{d}(x_n,x) \to 0$.


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