Proving a subspace (functions as vectors) Given the set
$U:=\{f:\mathbb{N}\rightarrow\mathbb{R}:f(n)=f(n-1)+f(n-2)$ for $n \geq 3\}$
I need to prove that $U$ is a sub space of
$V:=\{f:\mathbb{N}\rightarrow\mathbb{R}\}$

So basically I have to prove that things like $g+f$ and $\lambda f$ are again in $U$ (for some $f$, $g\in U$ and a $\lambda\in\mathbb{R}$)
How can I prove that?
 A: By doing it!  There is one possible complication.  Are you clear on the definition of "addition" and "scalar multiplication" for functions? 
 Given any functions, $f$ and $g$, how is $f+ g$ defined?  That is, if you know $f(x)$ and $g(x)$, how would you find $(f+ g)(x)$?
 Given a function, f, and a scalar, $\lambda$, how is $\lambda f$ defined?  How would you find $(\lambda f)(x)$?
Suppose f and g are both in that subspace.  Then $f(n)=f(n−1)+f(n−2)$ and $g(n)= g(n-1)+ g(n-2)$. So what is $(f+ g)(n)$?
Similarly, if f is in that subspace $f(n)= f(n-1)+ f(n-2)$.  For any scalar, $\lambda$, multiplying each side of that equation by $\lambda$, $\lambda f(n)= \lambda f(n-1)+ \lambda f(n-2)$.  But the definition of "scalar multiplication" for functions is precisely that $(\lambda f)(n)= \lambda f(n).
A: If $f,g \in U$ then
$$f(n)=f(n-1)+f(n-2)\\
g(n)=g(n-1)+g(n-2)$$
And if you call $h(n)=f(n)+g(n)$ then $h:\Bbb N\to \Bbb R$ and
$$h(n)=f(n-1)+f(n-2)+g(n-1)+g(n-2)=\\
h(n)=[f(n-1)+g(n-1)]+[f(n-2)+g(n-2)]=h(n-1)+h(n-2)$$
if $h(n)=\lambda f(n)$ then $h:\Bbb N\to \Bbb R$ and
$$h(n)=\lambda f(n-1)+\lambda f(n-2)=h(n-1)+h(n-2)$$
