## Help: isomorphism => completeness?

### Help: isomorphism => completeness?

If a normed space is isomorphic to $\mathbb R^n$, must it be complete?

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Help: isomorphism => completeness?

Yes. Isomorphism preserves all topological information.

In particular for completeness, you can see that isomorphism (and it's inverse) preserves distance (by definition) and limits (by continuity). So completeness is preserved automatically.

In fact, this is true for any topological space. A continuous bijective map between any two topological spaces preserve all toplogical information.

In particular for completeness, you can see that isomorphism (and it's inverse) preserves distance (by definition) and limits (by continuity). So completeness is preserved automatically.

In fact, this is true for any topological space. A continuous bijective map between any two topological spaces preserve all toplogical information.

### Re: Help: isomorphism => completeness?

By "preserves all topological information" I mean the two spaces are identical in every way, there is no difference between them except our notation.

### Re: Help: isomorphism => completeness?

But completeness is not a topological property?

To be specific, I can't conclude whether the space of all $n\times n$ matrices, with the operator norm, is complete or not. It is isomorphic to $\mathbb R^{n^2}$, but does this also mean that the norm is equivalent to any norm on $\mathbb R^{n^2}$?

To be specific, I can't conclude whether the space of all $n\times n$ matrices, with the operator norm, is complete or not. It is isomorphic to $\mathbb R^{n^2}$, but does this also mean that the norm is equivalent to any norm on $\mathbb R^{n^2}$?

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Help: isomorphism => completeness?

Now my question sounds trivial, because any limit point must also clearly be an $n\times n$ matrix(?) However, for the sake of rigour, does my last statement in the above post hold?

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Help: isomorphism => completeness?

I think we are talking about different types of isomorphism. What is the definition of your isomorphism?nayel wrote:But completeness is not a topological property?

To be specific, I can't conclude whether the space of all $n\times n$ matrices, with the operator norm, is complete or not. It is isomorphic to $\mathbb R^{n^2}$, but does this also mean that the norm is equivalent to any norm on $\mathbb R^{n^2}$?

### Re: Help: isomorphism => completeness?

A bijective linear map between vector spaces.

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Help: isomorphism => completeness?

Yes, then any finite dimensional vector space is complete. Because any finite dimensional vector space is isomorphic to $\mathbb{R}^n$ for some $n$. Any linear map between finite dimensional vector spaces is continuous. Therefore any bijective linear map is a homeomorphism (i.e. bijective and continuous). Therefore two isomorphic normed vector spaces are homeomorphic. So all finite dimensional vector spaces are homeomorphic to $\mathbb{R}^n$ for some $n$. Homeomorphism (bijective continuous function) preserves completeness. So all finite dimensional vector spaces are complete.

The easy way to see this is that the isomorphism and it's inverse are both bijective linear maps, therefore continuous. And continuous maps preserve limits. So completeness is also preserved.

The easy way to see this is that the isomorphism and it's inverse are both bijective linear maps, therefore continuous. And continuous maps preserve limits. So completeness is also preserved.

### Re: Help: isomorphism => completeness?

But homeomorphisms don't preserve completeness. For example, the spaces $\mathbb R$ and $(0,1)$ (with the usual metrics) are homeomorphic but the latter isn't complete.

"Everything should be made as simple as possible, but not simpler." - Albert Einstein

### Re: Help: isomorphism => completeness?

Hmm.. I messed up. Completeness in norm is not topological. Topological completeness is a weaker concept.

For the original question, you can prove that all norms on a finite dimensional vector spaces are equivalent. I did this when I took analysis but cannot remember properly. Say $|| \cdot ||$ is a norm then it is easy to show that (using triangle inequality) $ || \cdot || \leq C \times$(Euclidean norm) where $C$ is some constant. But to show the other direction i.e. $ || \cdot || \geq D \times$(Euclidean norm) where $D$ is some constant is more difficult.

For the original question, you can prove that all norms on a finite dimensional vector spaces are equivalent. I did this when I took analysis but cannot remember properly. Say $|| \cdot ||$ is a norm then it is easy to show that (using triangle inequality) $ || \cdot || \leq C \times$(Euclidean norm) where $C$ is some constant. But to show the other direction i.e. $ || \cdot || \geq D \times$(Euclidean norm) where $D$ is some constant is more difficult.