Isomorphism between R^inf and a proper subset of R^inf

[jojo12345]

I'm fairly certain the following is a vector space isomorphism $$\phi :\mathbb{R}^\infty\rightarrow\mathbb{R}^\infty$$ where the vector space is the space of infinite sequences of real numbers and phi is defined by $$\phi(a_1,a_2,...)=(0,a_1,a_2,...)$$. The mapping is linear and the inverse seems to be well defined. Is my logic flawed?

 

[Hurkyl]

the inverse seems to be well defined

What's the inverse of (1, 0, 0, 0, ...)?

(Why are you asking about the inverse anyways?)

 

[jojo12345]

You're right. As I wrote it, phi isn't an isomorphism. I'll try again. Define $$\phi :\mathbb{R}^\infty\rightarrow S$$ ,where $$S\subset \mathbb{R}^\infty$$ is the subspace containing all infinite sequences of reals of the form $$(0,a_1,a_2,...)$$, as: $$\phi(a_1,a_2,...)=(0,a_1,a_2,...)$$. Define the inverse function $$\phi^{-1}:S\rightarrow\mathbb{R}^\infty$$ as: $$\phi^{-1}(0,a_1,a_2,...)=(a_1,a_2,...)$$. Does this establish an isomorphism between $$\mathbb{R}^\infty$$ and $$S$$?

 

[Hurkyl]

Let's see... the things you need are
1. phi is well-defined
2. phi is linear
3. phi^-1 is well-defined
4. phi^-1 is linear
5. The composition of phi with phi^-1 is the identity
6. The composition of phi^-1 with phi is the identity

You've checked all of those, right? Then congratulations, you have an isomorphism!


Incidentally, you can prove a more general theorem: if phi is injective can you show it's an isomorphism onto its image? (And thus, all you need for your problem is an injection that is not surjective)

 

[jojo12345]

yes, it must be because it's surjective onto its image. Also, I think that 1,2,3,5,6 combined imply 4. Thanks a lot!