does the direct sum of constant sequences and null sequences gives convergent sequence Vector space

Question

Question: Let $S$ be the vector space of convergent sequences. Let $W$ be the vector subspace of constant sequences and $N$ be the vector space of null sequences. Then show that $S=W\oplus N$

Attempt:

If $S$ is the vector space of convergent sequences then say the sequence be $(x_n)\in S$ then

i)-$(x_n)=(x)+(x_n-x)$ where $(x)\in W$ and $(x_n-x)\in N$ and

ii)-its easy to show that constant sequence which is convergent to $0$ is the zero sequence which is identity of $S$.

Then by the Proposition 1.9 LINEAR ALGEBRA DONE RIGHT By Sheldon Axler we conclude that $S$ is direct sum of $W$ and $N$

and finally can some one say the name of the book from which this problem might have been picked up.

Thanks in Advance

Answer 1

You'll find this problem in "Linear Algebra : A Geometric Approach" by S. Kumaresan (page 42 Exercise 2.3.32). The given problem asks us to show the vector space of all real sequences $S$ to be the direct sum of two of its subspaces ($W$ and $N$). However, I think he meant the vector space of all real convergent sequences $C$.

Your proof is right. Take any convergent sequence $(x_n)$ and say its limit is $x$. Then the limit of $(y_n)$ where $y_n=x_n-x$ is $0$. We can write $(x_n)=(x)+(y_n)$ which proves that $C=W+N$. Do you see that this was the only way we could write $(x_n)$ as the sum of a constant and a null sequence? This observation leads us to the desired conclusion.