Proving Completeness of $(\mathbb{R},d)$

[Shoelace Thm.]

Homework Statement

Prove that $(\mathbb{R},d)$, $d = \frac{\mid x - y \mid}{1 + \mid x - y \mid}$ is a complete metric space.

Homework Equations

The Attempt at a Solution

If $d_u = \mid x - y \mid$, then I can prove this for the Cauchy sequences in $(\mathbb{R},d)$ that are also Cauchy in $(\mathbb{R},d_u)$. But there may be Cauchy sequences in $(\mathbb{R},d)$ that are not Cauchy in $(\mathbb{R},d_u)$.

 

[Dick]

Homework Statement

Prove that $(\mathbb{R},d)$, $d = \frac{\mid x - y \mid}{1 + \mid x - y \mid}$ is a complete metric space.

Homework Equations

The Attempt at a Solution

If $d_u = \mid x - y \mid$, then I can prove this for the Cauchy sequences in $(\mathbb{R},d)$ that are also Cauchy in $(\mathbb{R},d_u)$. But there may be Cauchy sequences in $(\mathbb{R},d)$ that are not Cauchy in $(\mathbb{R},d_u)$.

Can you give me an example of one? I don't really think there are any.

 

[Shoelace Thm.]

I couldn't find one either. How can I prove there are none i.e., there is one-to-one correspondence between the Cauchy sequences of each space?

 

[Dick]

I couldn't find one either. How can I prove there are none i.e., there is one-to-one correspondence between the Cauchy sequences of each space?

Suppose you could find constants m and M such that $md(x,y) \le d_u(x,y) \le Md(x,y)$? You would have trouble with the M part of that, but remember when you are dealing with Cauchy sequences you only have to worry about small values of |x-y|.

 

[Shoelace Thm.]

I don't quite understand what you are getting at, but I have that for n,m sufficiently large and $x_n$ Cauchy in (R,d), $\frac{\mid x_n - x_m \mid}{1 + \mid x_n - x_m \mid} < \epsilon \rightarrow \mid x_n - x_m \mid < \frac{1}{1 - \epsilon}$. So $x_n$ is Cauchy in (R,d_u).

 

[Shoelace Thm.]

Does this work?

 

[jbunniii]

I don't quite understand what you are getting at, but I have that for n,m sufficiently large and $x_n$ Cauchy in (R,d), $\frac{\mid x_n - x_m \mid}{1 + \mid x_n - x_m \mid} < \epsilon \rightarrow \mid x_n - x_m \mid < \frac{1}{1 - \epsilon}$. So $x_n$ is Cauchy in (R,d_u).

I don't see how that works, because $1/(1-\epsilon)$ does not approach 0 as $\epsilon \rightarrow 0$.

Suppose that
$$\frac{|x-y|}{1 + |x-y|} \leq \frac{1}{2}$$
Then we have
$$\begin{align}
|x-y| &= \frac{|x-y|}{1+|x-y|} \cdot (1 + |x-y|) \\
&= \frac{|x-y|}{1+|x-y|} + \frac{|x-y|}{1+|x-y|} \cdot |x-y| \\
&<= \frac{|x-y|}{1+|x-y|} + \frac{1}{2}|x-y| \\
\end{align}$$
So
$$|x-y| \leq 2\frac{|x-y|}{1 + |x-y|}$$
This should allow you to conclude that Cauchy in $d$ implies Cauchy in $d_u$.

By the way, did you already prove that $d$ is a metric?

 

[Dick]

Does this work?

junniiii as already done a fine job of pointing out why it doesn't work. My point was that you only have to worry around small values of ε. If |x-y|<1 then d and d_u are almost the same. They differ by a factor of at most 2.

 

[Shoelace Thm.]

Dick: Ok I understand now.

jbunniii: Yes I did prove d is a metric.