# What does $\mathbb{R}^\mathbb{R}$ mean?

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!

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

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

• 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 Nov 23, 2019 at 20:28
• @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. Nov 23, 2019 at 20:30
• So in other words: can I treat $\mathbb{R}^\mathbb{R}$ as $\mathbb{R}^\infty$? @Gabriele Cassese Nov 23, 2019 at 20:33
• @RafałSzypulski i edited my answer to address these topics. Briefly answering, no, since $\mathbb{R}^\infty$ is usually used meaning $\mathbb{R}^\mathbb{N}$ Nov 23, 2019 at 20:38
• You are a lifesaver. Thank you! Nov 23, 2019 at 20:40