About a family of functions that are linearly independent For an integer $k$, define a function $f_k :  \Bbb R\to \Bbb R$ to take any value on $(k,k+1)$(not all zero) and $0$ on $\Bbb R\setminus(k,k+1).$ Let $\mathcal F=\{f_k : k\in\mathbb Z\}$ be a family of these functions. Note that the functions in $\mathcal F$ are linearly independent.
Now, let $D=\{A_i : i\in \mathbb Z\}$ be collection of subsets of $\mathbb Z$ such that each $A_ i$ is infinite and $|A_i\cap A_j|<\infty$. Let $$B={\left\{\sum_{k\in A} f_k: A\in D \right\}}.$$

Question: Is B linearly independent?

To see that $B$ is linearly independent, by using the definition, we need to show if $$c_1\sum_{k\in A_1} f_k+c_2\sum_{k\in A_2} f_k+\cdots+c_n\sum_{k\in A_n} f_k=0,\tag {1}$$ then $c_1=c_2=\cdots=c_n=0.$ As in the assumption about elements in $D,$ we may have like $A_1\cap A_2=\{1,2,\dots, m\}.$ So, the first two terms can be written like that $$(c_1+c_2)(f_1+f_2+\cdots+f_m)+ c_1\sum_{k\in A_1\setminus(A_1\cap A_2)} f_k+c_2\sum_{k\in A_2\setminus(A_1\cap A_2)} f_k.$$ Here is where I got stuck. Should I consider cases for each $A_i$? I might be making it more difficult than it could be. Any help would be greatly appreciated.
 A: I will start with equation $(1)$ in OP. We have that $$c_1\sum_{k\in A_1} f_k+c_2\sum_{k\in A_2} f_k+\cdots+c_n\sum_{k\in A_n} f_k=0,\tag {1}$$ and this must be true for all $x\in\mathbb R.$ By way of contradiction, assume that $c_1\neq 0.$ By the definition of the set $D$, $A_1\cap A_i$ is finite for each $i=2,3,\dots,n,$ so we have that $${\left \lvert \, \bigcup_{i=2}^{n} (A_1\cap A_i) \, \right \rvert}<\infty$$ since it is finite union of finite sets. Now, choose $m \in A_1 \setminus \bigcup_{i=2}^{n} (A_1\cap A_i).$ This is possible since $A_1$ is infinite and $\bigcup_{i=2}^{n} (A_1\cap A_i)$ is finite. Further, choose $x^{*}\in\Bbb R$ such that $f_m(x^*)\neq 0.$ This is possible because $f_m : \mathbb R \to \mathbb R$ is not identically zero. Notice that all functions in $\mathcal F$ except $f_m$ will have a value of zero at $x^*,$ i.e., $$f_{k}(x^{*})=0 \text{ for all }f_k\in\mathcal F\setminus\{f_{m}\}.$$ By plugging in  $x^{*}$ to $(1),$ we have that $c_1 f_{m}(x^{*})=0$ which implies that $f_{m}(x^{*})=0$ --- a contradiction. So, we must have that $c_1 = 0.$
