Triangle inequality metric space

[beetle2]

Homework Statement

Let $(X,\theta)$ be a metric space. Take $K > 0$and define.

$\theta : X \cross X \rightarrow \real_{0}^{+}$, $(x,y)\rightarrow \frac{K\phi(x,y)}{1+K\phi(x,y)}$

Show that $(X,\theta)$ is a metric space.

Homework Equations

can someone please check my triangle inequality?

The Attempt at a Solution

$\phi(x,z) \leq \frac{K\phi(x,y)}{1+K\phi(x,y)}$

$\leq \mid \frac{K\phi(x,y)}{1+K\phi(x,y)}\mid + \mid \frac{K\phi(y,z)}{1+K\phi(y,z)}\mid$

$= \mid \frac{K\phi(x,y)}{1+K\phi(x,y)} + \frac{K\phi(y,z)}{1+K\phi(y,z)}\mid$

$=\phi(x,y)+\phi(y,z)$

 

[lanedance]

apologies if I'm missing something.. but what is $\phi(x,y)$?

and shouldn't you be trying to show the triangle inequality holds for $\theta(x,y)$?

 

[boneill3]

Yeah your right it is a typo the metric space should be $(X,\phi)$


so triangle inequality is

$\phi(x,y) \leq \frac{K\phi(x,)}{1+K\phi(x,y)}$
$\leq \mid \frac{K\phi(x,z)}{1+K\phi(x,z)}\mid + \mid \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$

$= \mid \frac{K\phi(x,z)}{1+K\phi(x,z)} + \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$

$=\phi(x,z)+\phi(z,y)$

 

[beetle2]

Is it ok now?

 

[Mark44]

Yeah your right it is a typo the metric space should be $(X,\phi)$


so triangle inequality is

$\phi(x,y) \leq \frac{K\phi(x,)}{1+K\phi(x,y)}$
$\leq \mid \frac{K\phi(x,z)}{1+K\phi(x,z)}\mid + \mid \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$

$= \mid \frac{K\phi(x,z)}{1+K\phi(x,z)} + \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$

$=\phi(x,z)+\phi(z,y)$

To show that $(X, \phi)$ is a metric space, you need to show:
1) $\phi(x, y) = 0~\text{iff}~x = y$
2) $\phi(x, y) = \phi(y, x)$
3) $\phi(x, y) + \phi(y, z) \geq \phi(x, z)$

Presumably you have already shown 1 and 2 and are working on 3 here (with some corrections).

$$\phi(x,y) = \frac{K\phi(x, y)}{1+K\phi(x, y)}$$
In the line above, it should be =, by how phi(x, y) is defined in post #1, although how it is defined is hazy, since phi(x, y) is defined in terms of itself.

How do you get to the next line? What's the justification here?
$$\leq \mid \frac{K\phi(x, z)}{1+K\phi(x, z)}\mid + \mid \frac{K\phi(z, y)}{1+K\phi(z, y)}\mid$$

 

[boneill3]

I hope this clears it up. Sorry I'm not to good at latex.


Let$(X,\phi)$be a metric space. Take $K > 0$and define.
$\theta : X \cross X \rightarrow \real_{0}^{+}$,$(x,y)\rightarrow\frac{K\phi(x,y)}{1+K\phi(x,y)}$




show that $(X,\theta)$is a metric

so the triangle inequality
$\theta(x,y) = \frac{K\phi(x,)}{1+K\phi(x,y)}$
$\leq \mid \frac{K\phi(x,z)}{1+K\phi(x,z)}\mid + \mid <br /> <br /> \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$
$= \mid \frac{K\phi(x,z)}{1+K\phi(x,z)} + <br /> <br /> \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid$

$=\phi(x,z)+\phi(z,y)$

My justification is because I was given that $(X,\phi)$ was a metric

space, we know that $\phi(x,y)$ is non negative. That in turn with $<br /> K > 0$ ensures that $\theta : X \cross X \rightarrow <br /> <br /> \real_{0}^{+}$ is a well defined non negative function.

 

[boneill3]

It suppose to be
show
$(X,\theta)$
is a mertic

 

[Mark44]

OK, I think I understand what you're trying to say, now. You are given that $(X, \phi)$ is a metric space (which means that $\phi$ is a metric).

There is another function $\theta$, where $\theta$: X x X --> [0, $\infty$), with K > 0 and
$$\theta(x, y) = \frac{K\phi(x, y)}{1 + K\phi(x, y)}$$

You are trying to show that $\theta$ is a metric (not mertic), which is the same as saying that $(X, \theta)$ is a metric space. (A metric space is a set together with a function that measures distance between elements of the set.) Part of the definition of a metric is that it is nonnegative, so you don't need the absolute values. The other parts of the definition I showed in post 5.

Since $\phi$ is a metric, it satisfies the triangle inequality. I see that you are trying to work this into your proof, but what you have doesn't look right to me. Certainly you can say that $\phi(x, y) \leq \phi(x, z) + \phi(z, y)$, but you also have $\phi(x, y)$ in the denominator, and you can't just substitute things in directly. Getting this right will require some more thought.

 

[lanedance]

i think Mark's onto it now, but i have to say the fact that there are 2 different posters & theta and phi have been interchanged (incorrectly as i read it now) in every post has made it pretty confusing...

 

[Mark44]

Amen to that, lanedance!

 

[boneill3]

but you also have $\phi(x, y)$ in the denominator, and you can't just substitute things in directly. Getting this right will require some more thought.

Sorry I'm getting lost here. Do I need to somehow get $\phi(x, y)$ out of the denominator?

 

[Mark44]

You can use the fact that $\phi$ is a metric to say this:
$$K\phi(x, y) \leq K\phi(x, z) + K\phi(z, y)$$

because $\phi$ satisfies the triangle inequality.

But you can't just say that
$$\frac{K\phi(x, y)}{1+K\phi(x, y)} \leq \frac{K\phi(x, z)}{1+K\phi(x, z)} + \frac{K\phi(z, y)}{1+K\phi(z, y)}$$

without some justification, particular regarding those denominators.

 

[boneill3]

But isn't the justification that if K >0 and
$$K\phi(x, y)$$ is a metric that's non negative

That it can't be > $$\frac{K\phi(x, z)}{1+K\phi(x, z)} + \frac{K\phi(z, y)}{1+K\phi(z, y)}$$?

You've got that the denominator = at least 1. So if x = z the function is 0.

 

[Mark44]

No, there's more to it than that. And it's not $K\phi(x, y)$ that is a metric - $\phi$ is a metric, so $\phi(x, y) \geq 0$ for any x and y in X.

It's easy enough to show that $\theta(x, y) \geq 0$ because of how $\theta(x, y)$ is defined.

Have you already proved that $\theta(x, y) = 0$ iff x = y? Have you already proved that $\theta(x, y) = \theta(y, x)$?

If so, then you still need to prove that $\theta(x, y) \leq \theta(x, z) + \theta(z, y)$. The expression you have in post #13 is equal to $\theta(x, z) + \theta(z, y)$. You need to show that this value is >= $\theta(x, y)$.

 

[Tedjn]

@boneill3: Can you explain in more detail what is confusing you? It took some work, but I've convinced myself that θ is a metric.