# COMPLETE metric space

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• Feb 1st 2010, 08:57 PM
kingwinner
COMPLETE metric space

Any help/hints is greatly appreciated!

[note: also under discussion in math links forum]
• Feb 1st 2010, 09:05 PM
Drexel28
Quote:

Originally Posted by kingwinner

Any help/hints is greatly appreciated!

[note: also under discussion in math links forum]

Hmm. I do believe that if $\displaystyle \left\{F_n\right\}$ is a decreasing sequence of closed sets in a complete metric space, then $\displaystyle \bigcap_{n\in\mathbb{N}}F_n\ne\varnothing$
• Feb 2nd 2010, 01:07 AM
Defunkt
Quote:

Originally Posted by kingwinner

Any help/hints is greatly appreciated!

[note: also under discussion in math links forum]

By decreasing, do you mean that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$ or $\displaystyle diam F_k \to 0$ or both?

If only the first, simply take the rays $\displaystyle F_n = [n, \infty) \subset \mathbb{R}$
• Feb 2nd 2010, 02:07 PM
kingwinner
Quote:

Originally Posted by Defunkt
By decreasing, do you mean that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$ or $\displaystyle diam F_k \to 0$ or both?

If only the first, simply take the rays $\displaystyle F_n = [n, \infty) \subset \mathbb{R}$

By decreasing, I mean ONLY that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$

And we have to find a decreasing sequence of "closed balls" in a COMPLETE metric space with empty intersection.

Can you please explain why your example $\displaystyle F_n = [n, \infty) \subset \mathbb{R}$ satifies all the requirements? Why is it a closed ball??? What metric are you using? And why is the metric space complete?

Thanks a lot!
• Feb 2nd 2010, 04:02 PM
Defunkt
Quote:

Originally Posted by kingwinner
By decreasing, I mean ONLY that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$

And we have to find a decreasing sequence of "closed balls" in a COMPLETE metric space with empty intersection.

Can you please explain why your example $\displaystyle F_n = [n, \infty) \subset \mathbb{R}$ satifies all the requirements? Why is it a closed ball??? What metric are you using? And why is the metric space complete?

Thanks a lot!

This is with the standard topology on $\displaystyle \mathbb{R}$, which is induced by the standard metric $\displaystyle d_1:\mathbb{R} \times \mathbb{R} \to \mathbb{R^+}, ~ d_1(x,y) = |y-x|$

I also assume you have shown that $\displaystyle \mathbb{R}$ is a complete metric space with this metric.

Indeed, I did not see the requirement for closed balls, sorry for that. The sequence I proposed not one of balls. The sets are closed because any convergent sequence in $\displaystyle [n,\infty)$ will be wholly included in some interval $\displaystyle [n,k] \subset [n, \infty)$ for some $\displaystyle k > n$, which is obviously a closed set, and therefore contains all of its limit points, thus $\displaystyle [n, \infty)$ contains all its limit points and is therefore closed.

Sorry again for the confusion.
• Feb 2nd 2010, 04:49 PM
kingwinner
But how can we construct an example of closed balls which satisify the all the requirements in the question?

Looking at hint 2, I agree that I_n ={n,n+1,n+2,...} is a decreasing sequence of sets, but I don't understand why they are "closed balls"...

thanks.
• Feb 3rd 2010, 07:36 PM
kingwinner
Actually, I am confused even to understand what hints 1 & 2 given in the question meant...
For hint 1: What does it mean to be a metric "on N"? How can {n>=k} ever be closed balls? This just doesn't make any sense to me...
For hint 2: I just don't understand what I_n ={n,n+1,n+2,...} is. How is I_n related to our problem?

Also, in this problem, what is the "metric space" that we're talking about? A metric space is a "set" together with a metric. The metric is given in hint 2, but what is that "set" in our problem?

Could somebody kindly explain these, please?
To you, maybe this question is simple. But to me right now, it seems impossible. :(
Any help is greatly appreciated!!
• Feb 5th 2010, 02:45 PM
kingwinner
OK, now I have more progress...

With the metric d as given in the hint,
And I've found a decreasing sequence of closed balls with empty intersection.

But with the given metric d(m,n)=∑1/(2^k) where the sum is from k=m to k=n-1, how can we prove that (N,d) with the metric d as defined above is a complete metric space?? (i.e. every Cauchy sequence in N converges (in N))

Any help is greatly appreciated!
• Feb 6th 2010, 03:36 PM
Nyrox
Unless I'm doing something wrong, the metric in Hint 2, as is, does not yield a complete metric space. As I get it, the metric is, given $\displaystyle m,n \in \mathbb N$:

$\displaystyle d(m,n)=\sum_{k=\min\{m,n\}}^{\max\{m-1,n-1\}}\frac{1}{2^k}$

right? So, consider the sequence of points $\displaystyle (x_n)=n$. With this metric, it's a Cauchy sequence. Indeed, let $\displaystyle \epsilon >0$, and choose $\displaystyle p\in \mathbb N$ such that $\displaystyle \frac{1}{2^{p-2}}<\epsilon$. Then, for any two points $\displaystyle m,n\in\mathbb N$ such that $\displaystyle m,n>p$, we have

$\displaystyle d(m,n)\leq d(p,m)+d(p,n) = \sum_{k=p}^{m-1}\frac{1}{2^k}+\sum_{k=p}^{n-1}\frac{1}{2^k} \leq 2\sum_{k=p}^{\infty}\frac{1}{2^k} =\frac{1}{2^{p-2}}<\epsilon$.

However, it does not converge to any point in $\displaystyle \mathbb N$. Suppose it did converge to a point $\displaystyle n_0\in\mathbb N$, and consider the ball $\displaystyle B$ centered at $\displaystyle n$ of radius $\displaystyle \frac{1}{2^{n_0+1}}$. It is immediate to check that $\displaystyle B\subseteq \{1,2,\cdots ,n_0\}$, which does not contain infinitely many terms of the sequence $\displaystyle (x_n)$. Thus $\displaystyle \mathbb N$ with this metric is not a complete metric space.

Unless, as I said, I've misunderstood something, or made a mistake somewhere along the way.
• Feb 6th 2010, 06:31 PM
kingwinner
Quote:

Originally Posted by Nyrox
Unless I'm doing something wrong, the metric in Hint 2, as is, does not yield a complete metric space. As I get it, the metric is, given $\displaystyle m,n \in \mathbb N$:

$\displaystyle d(m,n)=\sum_{k=\min\{m,n\}}^{\max\{m-1,n-1\}}\frac{1}{2^k}$

right? So, consider the sequence of points $\displaystyle (x_n)=n$. With this metric, it's a Cauchy sequence. Indeed, let $\displaystyle \epsilon >0$, and choose $\displaystyle p\in \mathbb N$ such that $\displaystyle \frac{1}{2^{p-2}}<\epsilon$. Then, for any two points $\displaystyle m,n\in\mathbb N$ such that $\displaystyle m,n>p$, we have

$\displaystyle d(m,n)\leq d(p,m)+d(p,n) = \sum_{k=p}^{m-1}\frac{1}{2^k}+\sum_{k=p}^{n-1}\frac{1}{2^k} \leq 2\sum_{k=p}^{\infty}\frac{1}{2^k} =\frac{1}{2^{p-2}}<\epsilon$.

However, it does not converge to any point in $\displaystyle \mathbb N$. Suppose it did converge to a point $\displaystyle n_0\in\mathbb N$, and consider the ball $\displaystyle B$ centered at $\displaystyle n$ of radius $\displaystyle \frac{1}{2^{n_0+1}}$. It is immediate to check that $\displaystyle B\subseteq \{1,2,\cdots ,n_0\}$, which does not contain infinitely many terms of the sequence $\displaystyle (x_n)$. Thus $\displaystyle \mathbb N$ with this metric is not a complete metric space.

Unless, as I said, I've misunderstood something, or made a mistake somewhere along the way.

You're right, indeed as you showed, the metric does not give a complete metric space.

So I doubt there is a problem with this exercise, and I just looked at the latest version of my textbook, the question is still there exactly as worded above, but hint 2 is taken away (hint 1 is still there). So probably the author realized that there was a typo/mistake in the metric d given in hint 2. So after all that hard work, at the end we actually find out that our counterexample doesn't work. :(
But since the question is still there, the author seems to suggest that a counterexample that satifies all the requirements (including completeness) definitely exist, but it looks like we have to modify or come up with ANOTHER metric d. How can we come up with a metric d that would work and give a complete metric space?

Any help is greatly appreciated! :)
• Feb 7th 2010, 10:56 AM
kingwinner
So the problem is that the metric given in the hint does not give a COMPLETE metric space...

If I define d to be

d(m,n)= 1/2 + ∑1/(2^k)
where the sum is from k=m to k=n-1

will this work or not?
• Feb 7th 2010, 11:32 AM
Nyrox
It will never work due to an argument similar to what Drexel said: in a compact space (in this case, the first ball), any intersection of non-empty nested closed sets is non-empty.
• Feb 7th 2010, 12:00 PM
kingwinner
Quote:

Originally Posted by Nyrox
It will never work due to an argument similar to what Drexel said: in a complete space (in this case, the first ball), any intersection of non-empty nested closed sets is non-empty.

It's non-empty when the radii of the closed balls go to 0, but here in our problem we only require it to be a decreasing sequence. In our question, decreasing means simply that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$.

Actually the theorem says: a metric space is complete iff every decreasing sequence of closed ball with radii going to zero has nonempty intersection.

So if the radii are not going to zero, we should be able to find an example of a decreasing sequence of closed balls in a complete metric space with empty intersection.. The problem is what metric to use...

Does anyone have any idea?
Thanks for any inputs!
• Feb 8th 2010, 03:15 PM
Drexel28
Quote:

Originally Posted by kingwinner
It's non-empty when the radii of the closed balls go to 0, but here in our problem we only require it to be a decreasing sequence. In our question, decreasing means simply that $\displaystyle F_{k+1} \subset F_k ~ \forall k \geq 1$.

Actually the theorem says: a metric space is complete iff every decreasing sequence of closed ball with radii going to zero has nonempty intersection.

So if the radii are not going to zero, we should be able to find an example of a decreasing sequence of closed balls in a complete metric space with empty intersection.. The problem is what metric to use...

Does anyone have any idea?
Thanks for any inputs!

If the radii of the sets does not approach zero then it must contain more elements than a sequence of sets whose radius approaches zero...so??
• Feb 8th 2010, 04:08 PM
Jose27
The argument won't work with closed balls by the same theorem you stated but it isn't true that in a complete metric space every decreasing sequence of closed sets has non-empty intersection: take $\displaystyle X_n=\{ x\in \mathbb{R} : x\geq n \}$
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