About the isomorphism of 2 infinite-dimensional vector spaces

[sanctifier]

Notations:
V denotes a vector space
A, B, C, D denote subspaces of V respectively
≈ denotes the isomorphic relationship of the left and right operand
dim(?) denotes the dimension of "?"

Question:
Find a vector space V and decompositions:
V = A ⊕ B = C ⊕ D
with A≈C but B and D are not isomorphic.

My opinion:
dim(V)=dim(A)+dim(B)=dim(C)+dim(D) and dim(A)=dim(C), but dim(B)≠dim(D) since V may not be finite-dimensional. It's an idea not an example, would you make a concrete example of V?

Thanks for any help!

 

[VKint]

(Let ~ denote isomorphism, + a direct sum, and <S> the span of the set S. Sorry, but Latex seems to be out of commission.)

I think this works:

Let V be the set of all infinite ordered-tuples of real numbers with only finitely many nonzero entries, i.e., the set of all infinite sequences that eventually terminate, such as {3,2,1,0,0,...}. Let e_i denote the sequence with a 1 in the i^th place and zeros elsewhere. Then B = {e₁,e₂,...} is a basis for V over R. Let N₁ = {e₁} and N₂ = {e₁,e₂}. Then V ~ V + <N₁> ~ V + <N₂> (I think). Clearly, <N₁> is not isomorphic to <N₂>, since their dimensions differ.

 

[quasar987]

VKint, I think the OP is looking for internal sums decompositions V = A ⊕ B = C ⊕ D.

But I think this slight variation on your idea works: take A:=<e_2,e_3,...>, B:=<e_1>, C:= <e_3,e_4,...>, D:=<e_1,e_2>.

 

[sanctifier]

VKint,quasar987

Thanks!