Шар в Банаховом пространстве

[AGu]

id в сообщении #258158 писал(а):

Цитата:

and $F$ is traditionally $\mathbb R$ or $\mathbb C$

By the way, why?

I guess, you've got an answer in the same message:

id в сообщении #258158 писал(а):

sure, such spaces aren't of much use.

However, there are theories of "normed" spaces with rather weird "scalars." Moreover, the values of a norm are sometimes allowed to be weird. For instance, there are lattice-normed spaces. Vector lattices are suitable in this case, since they possess a modulus $|{\cdot}|$ required in the equality $\|\lambda x\|=|\lambda|\|x\|$ mentioned by Профессор Снэйп.

[id]

Ah, now I see.
The norm is real-valued function, so all experiments with $Z_p$ are doomed to be a waste of effort.

Thanks.

[Circiter]

AGu в сообщении #258172 писал(а):

For instance, there are lattice-normed spaces. Vector lattices are suitable in this case

What about the Banach lattices? This is an example of complete normed vector lattices, thus such lattices are Banach spaces (indeed, we can study embeddings of Banach spaces into Banach lattices.)

[AGu]

Circiter в сообщении #258211 писал(а):

AGu в сообщении #258172 писал(а):

For instance, there are lattice-normed spaces. Vector lattices are suitable in this case

What about the Banach lattices?

What do you mean by "what about"?
A Banach lattice is a normed space in the usual sense
(the scalars are real and the norm is real-valued).

P.S. The topic seems to become an offtopic.

[Circiter]

2AGu

Цитата:

What do you mean by "what about"?

"What about" means "try this fresh idea."

Цитата:

A Banach lattice is a normed space in the usual sense

But additionally it's complete, so it's a Banach, isn't it?

P.S.: In the first place I keep in view the original problem on existence of a finite countable ball in a Banach space.

[AGu]

Circiter в сообщении #258222 писал(а):

"What about" means "try this fresh idea."

Hmm... What idea?

Circiter в сообщении #258222 писал(а):

Цитата:

A Banach lattice is a normed space in the usual sense

But additionally it's complete, so it's a Banach, isn't it?

Yes, it is. But a ball in a nonzero Banach lattice (over $\mathbb R$) cannot be finite, since a Banach lattice is a normed space (over $\mathbb R$).

[Circiter]

2AGu

Цитата:

But a ball in a nonzero Banach lattice (over $\mathbb R$) cannot be finite

It's not obvious statement (for me, at least,) can you prove it? How the norm can prohibit finiteness of subsets?

[AGu]

Circiter в сообщении #258226 писал(а):

Цитата:

But a ball in a nonzero Banach lattice (over $\mathbb R$) cannot be finite

It's not obvious statement (for me, at least,) can you prove it?

It was proven in my first message in this thread. A ball is convex and is thus at least continual. This holds true for every nonzero normed space and, in particular, for every nonzero Banach lattice.

[Solunac]

Thanks guys. This was really nice (nice talk). I appreciate it. Еще раз спасибо.

Может кто-то од вас посмотреть мои первые проблем (topic26320.html). Это один из моих проблем домашнего задания. Я до сих пор не решили ее, и до сегодняшнего утра он остался один...