# Homomorphic normed linear spaces

1. Nov 14, 2006

### 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 ?

2. Nov 14, 2006

### Hurkyl

Staff Emeritus
Did you menan to say "isomorphic"?

Have you given much thought to these questions?

3. Nov 14, 2006

### 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.

4. Nov 15, 2006

### quasar987

A continuous bijection with a continuous inverse is an homeomorphism.

5. Nov 15, 2006

### dextercioby

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

Daniel.

6. Nov 15, 2006

### 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.

7. Nov 15, 2006

### HallsofIvy

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

8. Nov 15, 2006

### StatusX

To show R^n is not isomorphic to R^m as a vector space for n$\neq$m, 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?

9. Nov 15, 2006

### 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.