Re: ? normed space and metric space



On Mon, 18 Feb 2008 08:28:44 -0500, "Cheng Cosine"
<acosine@xxxxxxxxxx> wrote:


"David C. Ullrich" <dullrich@xxxxxxxxxxx> wrote in message
news:thtir39ngom5fo2k9vth1pc3avuolerdgr@xxxxxxxxxx
On Mon, 18 Feb 2008 02:43:52 -0500, "Cheng Cosine"
<acosine@xxxxxxxxxx> wrote:
...
What are the differences between a normed space and a metric space?

I thought a metric space means a space that comes with the "distance"

being defined, and likewise a normed space with a norm being defined.

But isn't norm kind of distance? Any differences?

Any normed space is indeed a metrix space, but not conversely.

First, we only talk about norms on _vector spaces_, and not every
metric space is a vector space. But even for vector spaces there's
a big difference - there exist metrizable vector spaces which are not
normable.


Thanks you two so that I understand a little bit more :)

So from "large" to "small" (larger contains smaller):

normed space > metric space > inner product space

Correct?

No.

But in reading defs of the above spaces, I am confused with a complete
space

and a compact space. What are their differences?

??? Read the definitions more carefully - they're simply
not the same thing at all.

...
For example, let C be the space of continuous functions on R.
For f, g in C and n a positive integer define

d_n(f,g) = sup{|f(t) - g(t)| : |t| <= n}.

Define

d(f,g) = sum (2^{-n} d_n(f,g)/(1+d_n(f,g)).

Then d is a metric on C, but there is no norm on C which gives
the same topology.


Well, I still don't see why defining a metric d by the above leads to

non-existence of norm.

Also, per Elliot's post:

A metric space (S,d) is a set S and a function d:S^2 -> R with
d(x,y) = 0 iff x = y
d(x,y) = d(y,x)
d(x,z) <= d(x,y) + d(y,z)

But the def you used for your d(f,g) involves an integer n, doesn't

that mean that you need another set to place n? However, a metric,

according to Elliot's post, only involves 1 set. What's goning on then?

Thank you,
by Cheng Cosine
Feb/18/2k8 NC






David C. Ullrich
.