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

  • Thread starter Shoelace Thm.
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In summary, the conversation discusses how to prove that (\mathbb{R},d), d = \frac{\mid x - y \mid}{1 + \mid x - y \mid}, is a complete metric space. The participants discuss the possibility of Cauchy sequences in (\mathbb{R},d) that are not Cauchy in (\mathbb{R},d_u), but ultimately conclude that there is a one-to-one correspondence between the Cauchy sequences of each space and that d is a metric. They also mention that d and d_u are almost the same when |x-y| is small.
  • #1
Shoelace Thm.
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Homework Statement


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

Homework Equations


The Attempt at a Solution


If [itex] d_u = \mid x - y \mid [/itex], then I can prove this for the Cauchy sequences in [itex] (\mathbb{R},d) [/itex] that are also Cauchy in [itex] (\mathbb{R},d_u) [/itex]. But there may be Cauchy sequences in [itex] (\mathbb{R},d) [/itex] that are not Cauchy in [itex] (\mathbb{R},d_u) [/itex].
 
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  • #2
Shoelace Thm. said:

Homework Statement


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


Homework Equations





The Attempt at a Solution


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

Can you give me an example of one? I don't really think there are any.
 
  • #3
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?
 
  • #4
Shoelace Thm. said:
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|.
 
  • #5
I don't quite understand what you are getting at, but I have that for n,m sufficiently large and [itex] x_n [/itex] Cauchy in (R,d), [itex] \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} [/itex]. So [itex] x_n [/itex] is Cauchy in (R,d_u).
 
  • #6
Does this work?
 
  • #7
Shoelace Thm. said:
I don't quite understand what you are getting at, but I have that for n,m sufficiently large and [itex] x_n [/itex] Cauchy in (R,d), [itex] \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} [/itex]. So [itex] x_n [/itex] 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?
 
  • #8
Shoelace Thm. said:
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.
 
  • #9
Dick: Ok I understand now.

jbunniii: Yes I did prove d is a metric.
 

What is completeness of $(\mathbb{R},d)$?

Completeness of $(\mathbb{R},d)$ is a mathematical property that indicates whether or not a set of real numbers is "complete" in terms of a given metric, d. It essentially means that there are no "gaps" or "holes" in the set, and that it contains all of its limit points.

How is completeness of $(\mathbb{R},d)$ proven?

Proving completeness of $(\mathbb{R},d)$ involves showing that every Cauchy sequence in the set converges to a limit that is also in the set. This can be done using various mathematical techniques, such as the Bolzano-Weierstrass theorem or the monotone convergence theorem.

Why is completeness of $(\mathbb{R},d)$ important?

Completeness of $(\mathbb{R},d)$ is important because it allows us to make certain assumptions and conclusions about the set of real numbers. For example, it guarantees that continuous functions on the set will have certain properties, and it allows us to define concepts such as integrals and derivatives.

Can completeness of $(\mathbb{R},d)$ be proven for other sets besides the real numbers?

Yes, completeness can be proven for other sets besides the real numbers. In fact, it is a property that is often studied in other metric spaces, such as $(\mathbb{C},d)$ or $(\mathbb{Q},d)$.

Is completeness of $(\mathbb{R},d)$ equivalent to the least upper bound property?

No, completeness of $(\mathbb{R},d)$ is not equivalent to the least upper bound property. While they are related, the least upper bound property only guarantees the existence of a supremum for a set, whereas completeness requires that the set also contains all of its limit points.

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