Show that $S_0$ is a subspace of $S$ "Let $S$ denote the set of all infinite sequences of real numbers with scalar multiplication and addition defined by $a\{b_n\} = \{ab_n\}$ (where $a$ is a scalar) and $\{b_n\} + \{c_n\} = \{b_n +c_n\}$.  Let $S_0$ be the set of $\{b_n\}$ such that $b_n\rightarrow0$ as $n\to\infty$.  Show that $S_0$ is a subspace of $S$."
To prove that the set is a subspace, vectors in the subspace must be closed under addition and scalar multiplication.  Is this really as simple as every vector in $S_0$ obeying these two rules because they are part of a vector space (given) that obeys them?  Why is it significant that $b_n \to 0$ as $n \to \infty$?
 A: As you mentioned,
you have to prove that $S_0$ satisfies the following:


*

*$S_0$ contains the zero vector (i.e. the constant zero sequence $0,0,0,\ldots$);

*if $(a_n),(b_n)\in S_0$, then $(a_n)+(b_n)=(a_n+b_n)\in S_0$; and

*if $k\in\mathbb R$ and $(a_n)\in S_0$, then $k(a_n)=(ka_n)\in S_0$.


According to your definition,
$S_0$ is the set of sequences $(a_n)$ such that $a_n\to 0$,
thus,
we can rephrase the above three items as follows:


*

*the zero sequence $0,0,0,\ldots$ converges to zero;

*if $(a_n),(b_n)$ both converge to zero, then $(a_n+b_n)$ converges to zero; and

*if $k\in\mathbb R$ and $(a_n)$ converges to zero, then $k(a_n)=(ka_n)$ converges to zero.


Then,
you're free to use any facts you know about limits from calculus to demonstrate this claim.
A: I guess what you mean is if sum of two elements of $S$ obeys the rules, then so will $S_0$, you should look at the definition of subspace again. To prove subspace,for  two elements of subspace say $a$ and $b$, $a+b$ and $ca$ must be in $\textbf{subspace}$, being in the given vector space won't prove it. so you must show that for two sequences which are converging to $0$ their multiple and sum will also converge to $0$
