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## Homework Statement

we have the vector space (ℝ,ℝ^N,+) of all sequences in ℝ. if A={(x_n) [itex]\in[/itex] ℝ^N | only finitely many components x_j differ from 0}, show that A is a linear subspace of ℝ^N. With which other vector space is this subspace isomorphous?

## Homework Equations

## The Attempt at a Solution

For the first question I would prove three criteria:1. A [itex]!= [/itex] ∅

2. A [itex]\subset[/itex] ℝ^N

3. if r,s [itex]\in[/itex] ℝ and (x_n),(y_n)[itex]\in[/itex] ℝ^N.

1. take for instance the sequence (x_n), where x=0 for all n [itex]\in[/itex] N. clearly (x_n)[itex]\in[/itex] A. So A !=∅.

2. choose (x_n) in A. then (x_n) [itex]\in[/itex] ℝ^N. So, A[itex]\subset[/itex] ℝ^N.

3. choose r,s in ℝ and (x_n),(y_n) in A. because only finitely many components x_j (and also y_j) will differ from 0, I would expect that from certain n_0, all x_n=0 for n ≥ n_0. and for certain n_1, all y_n=0 for n≥n_1. Thus if we choose any n≥n_2=max{n_0,n_1}, we will find that both x_n=y_n=0 so that rx_n + sy_n=0 for all n≥n_2. this means that r(x_n) + s(y_n)[itex]\in[/itex] A.

On the second question, I would try to find a subspace with the same dimension. But first I would have to find a basis for (ℝ,A,+) to reveal how many dimensions it has. here is where I am stuck.

thanks