Re: Density regarding to a new norm

On Sat, 14 Apr 2007 15:20:15 EDT, Adrian Duma <ady@xxxxxxxxx> wrote:

On Fri, 13 Apr 2007 07:09:16 EDT, Adrian Duma
<ady@xxxxxxxxx> wrote:

Was there some reason you reposted my post without
any additions or comments?

At the very least you should learn to _quote_
quoted
text properly - your post _appears_ as though it
was something _you_ wrote.

On Thu, 12 Apr 2007 19:21:52 EDT, Adrian Duma
<ady@xxxxxxxxx> wrote:

On Wed, 11 Apr 2007 11:09:35 EDT, Adrian Duma
<ady@xxxxxxxxx> ,
using a keyboard with no Enter key, wrote:

Let X be a linear space, and let A be a linearly
independent subset of X.
Suppose that |X|=|A|=c (i.e., the cardinality of
the continuum). Is it true
that there exists a norm ||.|| on X such that A
is
a dense subset of (X,||.||) ?

Yes.

First extend A to a basis A union B.

Now say Y is a normed vector space with |Y| = c,
such
that Y has a countable dense subset (y_n).
Recursively
choose z_n in Y with ||z_n - y_|| < 1/n, such
that
z_n is not in the span of z_1, ... z_{n-1}.

So {z_n} is an independent dense subset of Y.
Extend
{z_n} to a basis for Y, and write that basis as
S union T, where {z_n} subset A, |S| = c and |T|
=
|B|.

Now the union of a bijection of A onto S and a
bijection of B onto T extends to an isomorphism
of X onto Y, and now the norm on Y gives you the
norm on X that you want.

Heh-heh.

Best regards,
Ady.


David C. Ullrich


************************

David C. Ullrich

RECALL THAT A IS A SUBSET OF X, WHILE {z_n} IS A
SUBSET OF Y!

You might think about your attitude, considering that
you're asking strangers to do your homework for you.

THE LINE "where {z_n} subset A" WAS OBVIOUSLY A TYPE
FOR "where {z_n} subset S"!


This was my comment, if you couldn't read it before.

If I couldn't _read_ it before? I couldn't read it
because it wasn't there. Again, you _really_ need
to reconsider your attitude.

Ady.


************************

David C. Ullrich

IMHO, I think that any stuff posted on the net
can be easily read by all interested people.

Huh? Of course that's true - what's your point?
Things which were never posted cannot be read,
even if someone _thinks_ they were posted.

There was no homework for anybody, and your construction is
the classical one (Beauzamy, Day, Lacey, ...).

Again, huh? If you knew this then why did you post
the question in the first place?

Besides the mathematics does not live from "obvious" typo's.
Now, deeply reconsideriny my attitude, I'm wondering if
your word "heh-heh" belongs to some ancient tribal culture,
since I couldn't find it into any classical English dictionary.

It was a little expression meaning that I was pleased
at having found a solution to a problem that I had
no idea how to do at first.

P.S. And, please, take a look at the Professor Anisiu's
generalisation of this question, because it is, indeed, a very deep one.

_Again_, what's your point? This somehow shows that you
didn't repost my first post with no comments, or what?

Best regards,
Ady


************************

David C. Ullrich
.

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