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The question is given below:

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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?

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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 $\mathbb{R}$

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  • $\begingroup$ 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. $\endgroup$ – hopefully Sep 11 at 18:42
  • $\begingroup$ but how we get from $c_{1} + ... + c_{n} = 0$ that $c_{i} = 0$ for every i? $\endgroup$ – hopefully Sep 11 at 18:46
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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.

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    $\begingroup$ @Dayton I did not say that. Linear combinations must be finite. So it doesn't span the space. $\endgroup$ – Henno Brandsma Sep 11 at 17:07
  • $\begingroup$ Sorry you are right. I have changed my vote. $\endgroup$ – Dayton Sep 11 at 17:09

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