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February 27th, 2017, 10:05 PM   #1
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Metric spaces ?

Hey! So I have a question; could you help me with it? If you have any idea, feel free to share it.

Let E, D be a metric spaces. If x1,x2,...,xn ∈ E, prove that

D(x1,xn) <= d(x1,x2) + d(x2,x3) +.....+ d(xn-1, xn)

Hope you could help.

Last edited by skipjack; February 28th, 2017 at 10:17 AM.
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February 27th, 2017, 10:24 PM   #2
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What is D(x1,xn)? Did you mean d? You already have a metric space named D. Which you don't seem to be using in your question beyond confusing it with the metric on E.

Can you clear this up?

Oh wait do you mean Let (E ,D) be a metric space? Well that's fine but then what is d?

For copy/paste math symbols I like Type mathematical symbols - online keyboard. They have ∈ for example.

If your question means what I think it does, correcting for the mistakes, this is a straightforward application of the triangle inequality.

Last edited by Maschke; February 27th, 2017 at 10:27 PM.
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February 28th, 2017, 12:55 AM   #3
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I will write to you the question again
(d does mean distance)

Let (E, d) be a metric space. If x1, x2, x3, ... , xn ∈ E,
prove that
d(x1,xn) <= d(x1,x2) + d(x2,x3)+ ...... + d(xn-1,xn)
Thanks from topsquark

Last edited by skipjack; February 28th, 2017 at 10:20 AM.
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February 28th, 2017, 06:56 AM   #4
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Quote:
Originally Posted by 001wf View Post
I will write to you the question again
(d does mean distance)

Let (E, d) be a metric space. If x1, x2, x3, ... , xn ∈ E,
prove that
d(x1,xn) <= d(x1,x2) + d(x2,x3)+ ...... + d(xn-1,xn)
Straightforward application of triangle inequality. Try it for n = 2 and then n = 3 and then n = 4 then see if you can do the general case.

Last edited by skipjack; February 28th, 2017 at 10:20 AM.
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February 28th, 2017, 09:14 AM   #5
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Given: Space (set) E with a metric (distance function) D.

The triangle inequality is a postulate of D.

If you want to test whether a given function is a metric, that's a different problem. First you have to specify the set and the function.
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March 6th, 2017, 10:27 AM   #6
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Am really not English, i don't quite understand what you mean?
But could you solve it for me?
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March 6th, 2017, 10:38 AM   #7
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Quote:
Originally Posted by 001wf View Post
Am really not English, i don't quite understand what you mean?
But could you solve it for me?
If you are studying metric spaces, you have learned the triangle inequality, yes?

https://en.wikipedia.org/wiki/Triang...y#Metric_space
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March 6th, 2017, 05:40 PM   #8
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Ok I'll do this one for you.

We wish to prove that if $\{x_i\}_{i=1}^n$ is a set of points in a metric space, then $\displaystyle d(x_1, x_i) \leq \sum_{i=2}^n d(x_{i-1}, x_i)$.

We'll proceed by induction on the number of points. The base case is three points. If $x_1, x_2, x_3 \in E$, then $d(x_1, x_3) \leq d(x_1, x_2) + d(x_2, x_3)$. This is the famous triangle inequality. It's true for any metric. In fact it's one of the defining properties of a metric.

The induction hypothesis is that this works for $n - 1$ points; in other words $\displaystyle d(x_1, x_{n-1}) \leq \sum_{i=2}^{n-1} d(x_{i-1}, x_i)$.

Now that we have the base case and the induction hypothesis, suppose we have $n$ points. Then

$d(x_1, x_n) \leq d(x_1, x_{n-1}) + d(x_{n-1}, x_n) ~~~~~ $ [by plain old triangle inequality]

$\displaystyle \leq \sum_{i=2}^{n-1} d(x_{i-1}, x_i) + d(x_{n-1}, x_n) ~~~~~ $ [first summand is by induction hypothesis]

$\displaystyle = \sum_{i=2}^n d(x_{i-1}, x_i)$

which is what we are required to prove. $\square$

I want to mention that I did not just type all this notation in. I had to work at it with pencil and paper to make sure all the subscripts and summation indices were correct and expressing the right idea. I didn't want you to think you are supposed to see all this right away. We all have to work at this kind of detail.

As far as knowing to use induction, that's a matter of experience. You know the triangle inequality works for three points, and you are asked to prove it works for $n$ points. Induction is often helpful in a situation like that.

And as far as how you would know that the triangle inequality was the key to the problem, you'll find in metric spaces that the triangle inequality is the key to pretty much every problem. Because in a metric space, the triangle inequality is the only interesting thing we know.

Last edited by Maschke; March 6th, 2017 at 05:49 PM.
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