Triangle inequality metric space

[beetle2]

Homework Statement

Let (X,\theta) be a metric space. Take K > 0and 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 > 0and 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 /> <br /> K &gt; 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.