Are the geometric sequences a subspace of all infinite sequences of real numbers

Question

I'm a little confused about how to prove that the given infinite sequence is a subset of "the space of all infinite sequences."

so V = subspace of all infinite sequence of real numbers so $$(x_0, x_1,...)$$ where they add element by element so $$(x_0, x_1...) + (y_0, y_1, ...) = (x_0+y_0, y_1+y_2, ...)$$

G = geometric sequences, so sequences of the form $$(a, ar, ar^2, ar^3, ...)$$ A = arithmetic sequences, so sequences of the form $$(a,a+k,a+2k,a+3k, ...)$$

So, I know we need to show three things (1) $0$ is in the subspace, (2) the subspace is closed under addition, and (3) closed under multiplication.

1. for any $a = 0$ and $k = 0$ then $0 \in A$, and for any $a = 0$ or $r = 0$ then $0 \in G$

2. for any $p,q \in A$ then $(p+q)\in A$ (then the same for $G$)

Here's where I get confused, If I pick $p = (a, a+k, a+2k,...)$ and $q = (a, a+k, a+2k, ..)$, then $$(p+q) = (2a,2a+2k,2a+4k,...)$$, and since $a$ and $k$ are just constants we replace $a=2a$ and $k=2k$, then this is of the form that we want. We then do pretty much the same thing with $\gamma(a,a+2k,a+3k,...)$ and then we get the answer we want (for A) that it is a subspace.

However, if we do the same thing for $G$, then we get the same answer that it is a subspace (and the answer key say it is not).

I think the issues lies in picking $p = q$ and assuming we can replace $a=2a, k=2k$, however I don't see how to pick $p$ and $q$ differently such that we can show $A$ is a subspace and $G$ is not a subspace. Can someone point out what I'm misunderstanding/doing incorrectly. Thanks!

Answer 1

The point is that the sum of any two elements of the 'sunspace' must be an element, so you cannot assume $p=q$. Try it with $p=a,ar,ar^2\dots ,q=b.bs,bs^2,\dots$ and it fails. For the corresponding thing on arithmetic sequences, it still works.

Answer 2

What you need to decide is whether $a+b$, with $a=(a,ar,..,ar^n,....)+b=(b,br',..,br'^n,...)$ (pointwise-addition) is a geometric series.

We get $$ a+b=(a+b, ar+br',.....,ar^n+br'^n,........)$$

For $a+b$ to be a geometric series, we need the ratio of consecutive terms to be constant, so that , $ \frac{ ar^{n+1}+br'^{n+1}}{ar^n+br'n}=k. ,k$ a constant, for all Reals $r,r'$. It is straightforward to find a counterexample: Try, e.g., fixing $r$( or $r'$) to be $1$, and choose any other value for $r'$( or $r$).