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Im having problem with the following task:
Let $ V = \mathbb{R}^\mathbb{R}$, $ \mathbb{K} = \mathbb{R}$ and $A = (\sin(x),\sin(2x),\sin(3x),\sin(4x))$.
Is $A$ linearly independent? Does $V = {\rm Lin}(A)$? Is $A$ the basis of $V$ over $K$?
The first part is pretty easy. All I had to do was to use Wronskian determinant, but how should I solve the other two parts? And what does $\mathbb{R}^\mathbb{R}$ mean? Is this equivalent to $\mathbb{R}^\infty$?
Any help is appreciated. Thanks in advance!
It might be a set of all functions from R to R.
It is the set of all the function $\mathbb{R}\to \mathbb{R}$. This is true in general for every $X^Y$ (as noted in the comments by Lord Shark the Unknown), and somewhat follows from $X^Y:=\Pi_{Y}X$ (the idea here is that one of the possible equivalent definition of a cartesian product $\Pi_{\mathcal{F}}X_f$ is the family of function from that associates $f\in \mathcal{F}$ with some element of $X_{f}$. The existence of such a function, called choice function, is highly non trivial and follows from the axiom of choice as you can see here).
Now, I claim that $\mathbb{R}^\mathbb{R}$ is not finitely generated. To prove it, let $\mathbb{P}$ be the vector subspace of all the polynomials. Now, given any finite l.i. set of vectors in $\mathbb{P}$, their products cannot be obtained as a linear combination of those vectors. Thus, $\mathbb{P}$ is not finitely generated, and so $\mathbb{R}^\mathbb{R}\supset \mathbb{P}$ cannot be finitely generated either.
Thus, $A$ is not a basis of $\mathbb{R}^\mathbb{R}$, as neither is any finite set of vectors.
Note: $\mathbb{R}^\infty$ is usually used to denote $\mathbb{R}^\mathbb{N}$, and so is not an equivalent notation
