I am not that familiar with formatting on here, so I placed the question here:
I hope that is okay.
Here is my attempt at the question.
a)
We need to show that W is closed under addition and scalar multiplication. Let f, g be elements of W, and c be an element of Real numbers.
Addition
We need to show that f+g have a support which is finite. Choose this to be the set of values n (call this set U) which belong to the union of the finite sets for which supp f(n) is finite (call this set S) and supp g(n) is finite (call this set G). Then (f+g)(n) = f(n) + g(n) = 0 for all n which are not an element of U. Therefore, it is closed under addition.
Scalar Multiplication
Then (cf)(n) = cf(n) = 0 for all n which are not an element of S, since f(n) = 0, and zero multiplied by any scalar is 0. Therefore it is closed under multiplication.
b)
It is linearly independent and spans W, so it is a basis.
Span:
Any function of W can be expressed as a linear combination of fn. Because a function of W has a finite support, then there is a set of values x, for which f(x) is non zero. Denote those x as x1, x2, ...xn. Then (f(x1), f(x2), ...f(xn)) can be expressed as (c1fx1(x1), c2fx2(x2)...cnfxn(xn))
Linear Independence.
Help on this would be appreciated.
For (b), you need to first understand what the zero vector is in $W$. It happens to be $f(x)=0$, i.e. no support. Call this function $Z(x)$.
Now, suppose that $\{f_n\}$ were linearly dependent. Then there would be a finite set $S\subseteq \mathbb{Z}$ and some constants $a_n$ with $$Z(x)=\sum_{n\in S} a_nf_n(x)$$
Now, for each $m\in S$, we have $$0=Z(m)=\sum_{n\in S}a_nf_n(m)=a_mf_m(m)=a_m$$ Hence in fact all the constants were $0$.