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!

  • 4
    $\begingroup$ If $X$ and $Y$ are sets, then $X^Y$ is the set of functions from $Y$ to $X$. $\endgroup$ Nov 23, 2019 at 20:16

2 Answers 2


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

  • $\begingroup$ Thanks you for your answer! :) Any ideas on how I should solve the second part? Is the statement that since dim(R over R) is 1 then A isnt a base of V over K but it does span it correct? @Gabriele Cassese $\endgroup$ Nov 23, 2019 at 20:28
  • $\begingroup$ @RafałSzypulski $A$ is not a base of $V$ over $K$. Actually, no finite set of linearly indipendent vectors from $V$ is a base of $V$, since $V$ is not finite dimensional. $\endgroup$
    – Caffeine
    Nov 23, 2019 at 20:30
  • $\begingroup$ So in other words: can I treat $\mathbb{R}^\mathbb{R}$ as $\mathbb{R}^\infty$? @Gabriele Cassese $\endgroup$ Nov 23, 2019 at 20:33
  • $\begingroup$ @RafałSzypulski i edited my answer to address these topics. Briefly answering, no, since $\mathbb{R}^\infty$ is usually used meaning $\mathbb{R}^\mathbb{N}$ $\endgroup$
    – Caffeine
    Nov 23, 2019 at 20:38
  • $\begingroup$ You are a lifesaver. Thank you! $\endgroup$ Nov 23, 2019 at 20:40

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