Homomorphic normed linear spaces

[tuananh]

Hi all experts,
I've just visitted another Maths forum and picked up two interesting questions:
Question 1. Let H and K be homomorphic normed linear spaces. Is it necessary that H and K have the same dimension if both H and K are finite-dimension ? Is there possible a homomorphism between an infinite-dimension normed space and a finite-dimension one ?
Question 2. If f is a homomorphism between two normed linear spaces, is f necessary uniformly continuous ?
Hope to get your ideas.

 

[Hurkyl]

Let H and K be homomorphic normed linear spaces.

Did you menan to say "isomorphic"?


Have you given much thought to these questions?

 

[tuananh]

By saying "H and K is homomorphic", I mean that there is an one-to-one mapping f from H to K such that both f and f^{-1} (the inverse of f) are continous. Perhaps, this concept should be called "isomorphic" and sorry if I made a mistake.

If H=R and K=R^2, I think R and R^2 cannot be holomorphic because if we remove a point of R, such as x, then R\{x} is no longer a connected set, but R^2\{f(x)} is still a connected set. Here, f is an one-to-one from R to R^2 such that f and f^{-1} are continuos.

 

[quasar987]

A continuous bijection with a continuous inverse is an homeomorphism.

 

[dextercioby]

A one-to-one homomorphism between 2 algebraic structures is an isomorphism.


Daniel.

 

[matt grime]

No it isn't. The inclusion of Z into Q or R (as additive groups) is a one to one homomorphism. It is not an isomorphism.

 

[HallsofIvy]

So, I think the phrase needed here is "homeomorphic isomorphism" or perhaps "isomorphic homeomorphism"!

 

[StatusX]

To show R^n is not isomorphic to R^m as a vector space for n\neqm, you can just show that isomorphism preserves dimension (ie, show a basis is mapped to a basis). Showing they are not homeomorphic as topological spaces is much more difficult. It can be done using cut points as in post 3 for n=1, but above this you need more advanced tools from algebraic topology. And what do you mean by uniformly continuous?

 

[mathwonk]

real normed linear spaces of finite dimension are probably always isomorphic to R^n. then they are determined by their homeomorphism type i would guess. a proof tends to involve homology theory.