# Is $\{f_{a} \mid a \in \mathbb{R}\}$ a basis for $\mathbb{R}^{\mathbb{R}}$.

The question is given below:

I know the definition of the basis, it must be linearly independent and it must span the space, but I am stucked in applying the definition here, could anyone help me please in doing so?

To check linear independence we have to check that every finite subset of $$\{f_a\mid a\in\mathbb{R}\}$$ is linearly independent. To this end let $$\{a_1,\dotsc, a_n\}$$ be a collection of distinct real numbers and suppose that $$c_1f_{a_1}+\dotsb+c_nf_{a_n}=0$$ for some $$c_i\in\mathbb{R}$$ where the $$0$$ denotes the zero function. Evaluate both sides at $$a_i$$ to conclude that $$c_i=0$$.
For spanning, note that the span of $$\{f_a\mid a\in\mathbb{R}\}$$ equals the set of functions $$E$$ that are equal to zero except at finitely many points. In particular $$f(x)=x$$ is not in the span and $$E\ne \mathbb{R}^{\mathbb{R}}$$ whence $$\{f_a\mid a\in\mathbb{R}\}$$ is not a basis for $$\mathbb{R}^{\mathbb{R}}$$ over $$\mathbb{R}$$
• Could you please include more details for the evaluation at $a_{i}$? I understand the idea but I am confused about the correct way of writing it i.e. I know that $f_{a_{1}} (a_{i})$ = 1 if i=1 and zero otherwise. – hopefully Sep 11 at 18:42
• but how we get from $c_{1} + ... + c_{n} = 0$ that $c_{i} = 0$ for every i? – hopefully Sep 11 at 18:46
No, it is not. It is a linearly independent set but you cannot write $$f(x)=x$$ as a finite linear combination of $$f_a$$'s.