# My Topology Questions

1. Nov 16, 2006

### JasonRox

Alright, instead of starting a new thread everytime I have a question, I will just post it in here.

Note: These are not from assignments.

Note: Most of these questions can be found in Topology by Munkres. I will make a mention when it is, and where it is.

So, here is the first one...

This is from Munkres, found on page 126.

Define the following metric on R^n as follows...

$$d'(x,y) = \sum_{i=1}^{n} \left[ |x_i - y_i|^{p} \right]^{1/p}$$

...where p >= 1.

Show that it induces the usual topology on R^n.

I barely know the first step to showing that if x is in d-ball (usual R^n ball), then there exists a d'-ball that contains x and it is contained in d-ball.

I'll spend the next few minutes thinking about it, or longer. I would certainly ask my prof. tomorrow.

Last edited: Nov 16, 2006
2. Nov 16, 2006

### AKG

Draw some pictures of the d'-balls to get some insight into what d-balls contain and are contained in a d'-ball.

3. Nov 17, 2006

### JasonRox

Actually, I do know what it looks like.

My problem is writing a proof because I know I need to include the fact that p is bigger than 1, but I guess I don't know where that comes into play.

4. Nov 17, 2006

### matt grime

It's nothing to do with topology. That's the first hint.

The second, if you scrolled down this far, is that it is all to do with analytical results you know (I think).

The third is, that this result is often proven in probability theory - it is something like Jensen's inequality, or the fact that the r'th moment is larger than the r-1'st moment, or something to do with concave functions etc.

5. Nov 17, 2006

### AKG

I think you mean to put the "1/p" around the whole sum, not on each term. Anyways, if you draw pictures, you get a sense of where the boundary of a d'-ball is farthest from its center and where it's closest (it changes for p < 2 and p > 2 I think). In other words, if you have a d'-ball centered at x, you can give neat expressions for the 2n points where the boundary is closest to x and the 2n points where the boundary is furthest. Take one of those closest points, and find its d-distance from x. Then a d-ball of that radius is inside your d'-ball. Take one of the farthest points and find its d-distance from x. Then a d-ball of that radius contains your d'-ball.

6. Nov 18, 2006

### matt grime

You might also want to look through some functional analysis texts, since you are showing that all norms on a finite dimensional banach space are equivalent; this is one of the first things shown and should be easy to find.

7. Nov 18, 2006

### JasonRox

This is exactly what I was going to do.

It's what I thought of yesterday or the night I wrote this thread.

I'll just have to get the explicit representation, which I'll have to wait until Monday.

Note: Yes, the "1/p" goes around the whole sum.

Last edited: Nov 18, 2006
8. Nov 18, 2006

### JasonRox

I knew it had nothing to do with topology.

Thanks for the mention in regards to probability theory. I'll look into it for sure. It's probably in my text somewhere.

9. Nov 18, 2006

### JasonRox

Great!

Two nice references to where this is done.

Once I associate this problem with other things, I'm most likely to not forget it, which is a bonus.

10. Nov 19, 2006

### matt grime

I should have made it clear that there are two different things I've mentioned.

Functional analysis will lead you to the conclusion that the norms are equivalent: to show two metrics on a finite dimensional vector space are equivalent it suffices to shwo this if one is the standard metric. Metrics are continuous function in the metric topology, tautologically, and you use some results like continuous functions on compact spaces attain their max and min values, it is elegant and short, though I can't recall the precise argument at the moment.

The probability theory avenue will lead you to a stronger result, that the ball of radius e in the p metric is contained in the ball of radius e in the q metric for q=>p. Actually the result I'm thinking of states that
E(|x-x'|^p)^{1/p} <= E(|x-x'|^p+1)^{1/p+1}

E is the expectation operator and x' is the mean of x. But the proof, using Jensen's inequality actually proves the stronger result if interpreted correctly.

Last edited: Nov 19, 2006
11. Nov 19, 2006

### mathwonk

in a finite dimensional normed space, the unit ball is compact, hence the identity map restricted to it has an image if minimum positive norm. that proves that any two norms are equivalent.