# 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.

• Exactly what do you mean when you say that each element of $B$ is linearly independent? Commented May 26, 2021 at 14:55
• @Carlo, it was a mistake. I have in mind that the family $\mathcal F$. I fixed
– 00GB
Commented May 26, 2021 at 15:01
• I am not sure how to deal with the sum you have written down. It is a finite linear combination, but it is an infinite sum (because each $A_i$ is infinite). Commented May 26, 2021 at 15:13
• @Carlo, I think i have some ideas and I will write them soon
– 00GB
Commented May 26, 2021 at 15:16

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 $$c_1 = 0.$$