Verifying a Vector Space Via Given Axioms Let $X$ be the collection of all sequences $\{\alpha_n\}_{n=1}^{\infty}$ of scalars from $\mathbb{K}$ such that $\alpha_n=0$ for all but a finite number of values of $n$. Define addition and scalar multiplication on $X$ by
$\{\alpha_n\} + \{\beta_n\} = \{\alpha_n + \beta_n \}$ and $\lambda \{\alpha_n\} = \{\lambda \alpha_n\}$. 
Verify that $X$ is a vector space over $\mathbb{K}$. 
According to my textbook which is being used for this problem, I need to verify the property of closure under addition and scalar multiplication as well as the following 8 axioms:
1). $x+(y+z)=(x+y)+z$
$\{\alpha_n\} + (\{\beta_n + \gamma_n\}) = \{\alpha_n + \beta_n + \gamma_n\}$ and similarly, $\{(\alpha_n + \beta_n)\} + \{\gamma_n\} = \{\alpha_n + \beta_n + \gamma_n\}$ by definition of addition on $X$.
2).$x+y=y+x$
$\{\alpha_n\} + \{\beta_n\} = \{\beta_n\} + \{\alpha_n\}$.
Since $\{\alpha_n\} = 0$ for all but finitely many values of $n$, then $\{\alpha_n\} + \{\beta_n\} = 0+0 = \{\beta_n\} + \{\alpha_n\} = 0+0$.
3). $x+0 = x$.
$\{\alpha_n\} + 0 = \{\alpha_n\}$.
By the definition of addition on $X, \{\alpha_n\} + 0 = \{\alpha_n + 0\} = \{\alpha_n\}$.
4). $x+ (-1)x= 0$. 
$\{\alpha_n\} + (-1)\{\alpha_n\}=0$.
Let $\lambda = -1$ and we use the definition of scalar multiplication: 
$\{\alpha_n\}+ \lambda \{\alpha_n\} = \{\alpha_n\} + \{\lambda \alpha_n\}$. Using the definition of addition on $X$, we get $\{\alpha_n+ \lambda \alpha_n\} = \{\alpha_n - \alpha_n\}=0$.
5). $(\lambda + \mu)\{\alpha_n\} = \lambda \{\alpha_n\} + \mu \{\alpha_n\}$.
Using the definition of scalar multiplication on $X$, we can write $(\lambda+\mu)\{\alpha_n\} = \{(\lambda + \mu)\alpha_n\} = \{\lambda \alpha_n + \mu \alpha_n\}$. Using addition on $X$, we can have $\{\lambda \alpha_n + \mu \alpha_n\} = \{\lambda \alpha_n\} + \{\mu \alpha_n\}$. Again using scalar multiplication, we have $\lambda \{\alpha_n\} + \mu \{\alpha_n\}$.
6). $\lambda \{\alpha_n + \beta_n\} = \lambda \{\alpha_n\}+ \lambda\{\beta_n\}$.
By scalar multiplication on $X$,  $\lambda \{\alpha_n + \beta_n\} = \{\lambda(\alpha_n + \beta_n)\}$. Using addition, $\lambda\{\alpha_n+\beta_n\} = \lambda[\{\alpha_n\}+\{\beta_n\}]$. By scalar multiplication, $\lambda[\{\alpha_n\}+\{\beta_n\}] = \{\lambda \alpha_n\} + \{\lambda \beta_n\}$. Again by scalar multiplication, $\lambda\{ \alpha_n\} +\lambda \{ \beta_n\}$.
7). $(\lambda \mu)\{\alpha_n\} = \lambda (\mu \{\alpha_n\})$.
By scalar multiplication, $(\lambda \mu)\{\alpha_n\} = \{(\lambda \mu)\alpha_n\}$. Writing $(\lambda \mu) = (\lambda)(\mu)$, we have $(\lambda)(\mu)\{\alpha_n\}$. By scalar multiplication, $(\lambda)(\mu)\{\alpha_n\} = \lambda\{\mu \alpha_n\}$.
8). $1 \cdot \{\alpha_n\} = \{\alpha_n\}$.
Let $\lambda=1$. By scalar multiplication, $\lambda \{\alpha_n\} = \{\lambda \alpha_n\} = \{1 \cdot \alpha_n\} = \{\alpha_n\}$.
I have not shown the property of closure under addition and scalar multiplication since I am unsure about how to do that with my given information. Any help/advice/suggestions will be greatly appreciated. Thanks in advance.
The textbook I am using is Functional Analysis An Elementary Introduction by Markus Haase.
 A: Well, I think that by "closure" your text-book means that addition and scalar multiplication are well-defined. Indeed, call $V$ your candidate vector space of all the sequences with finite support (i.e., with a finite number of non-zero terms), then you can think to $V$ as a subset of the cartesian product $\mathbb K^{\mathbb N}$ (the space of all sequences indexed by $\mathbb N$ in your field). Of course addition is defined as a function
$$\mathbb K^{\mathbb N}\times \mathbb K^{\mathbb N}\to \mathbb K^{\mathbb N}$$
and similarly scalar multiplication is defined as 
$$\mathbb K^{\mathbb N}\times \mathbb K\to \mathbb K^{\mathbb N}\,.$$
Now, if you restrict the domain respectively from $\mathbb K^{\mathbb N}\times \mathbb K^{\mathbb N}$ to $V\times V$ and from $\mathbb K^{\mathbb N}\times \mathbb K$ to $V\times \mathbb K$, you have to check that the codomain is in $V$. In other words, the sum of two sequences with finite support, as well as a multiple of such a sequence, has again finite support. For this you have just to use the elementary property that $0+0=0$ and $0\cdot\lambda=0$ for all $\lambda \in \mathbb K$.
EDIT: Just to be completely explicit, take two elements $\{a_n\}_n\,, \{b_n\}_n$ in $V$ and $\lambda\in \mathbb K$. Then, there exists $m\in\mathbb N$ such that $a_n=b_n=0$ for all $n\geq m$. Of course then $a_n+b_n=0$ and $a_n\lambda=0$ for all $n\geq m$, showing that both $\{a_n+b_n\}_n$ and $\{a_n\}_n\lambda$ belong to $V$. 
