I’d really some help on this problem i’ve been working on for while. I’ve got two sets $X$ and $Y$, the map $overline{phi}:mathbb{R}^X times mathbb{R}^Y rightarrow mathbb{R}^{Xtimes Y}$ such that $overline{phi}(f,g)=f(x)times g(y)$. $overline{phi}$ is clearly bilinear under the standar definition of function adition and scalar multiplication, so there exists a linear map $psi :mathbb{R}^X otimes mathbb{R}^Y rightarrow mathbb{R}^{X times Y}$ such that $overline{phi}(f,g)=psi (fotimes g)$.

I need to prove $psi$ is onto iff $X$ or $Y$ are finite.

So far i have tried defining a basis for

$mathbb{R}^X times mathbb{R}^Y$ as:

begin{equation}

notag

{(delta_{x_1},0),dots, (delta_{x_{|x|}},0), (0,delta_{y_1}),dots , (0, delta_{y_{|Y|}}) }

end{equation}

Where $delta_{x_i}(x)=1 text{if} x_i=x$, otherwise it’s zero. Thus, a basis for the tensor product will be ${ (delta_{x_i}otimes delta_{y_j})|x_iin X, y_jin Y }$. A basis for $mathbb{R}^{Xtimes Y}$ is ${ delta_{x_i,y_j}(x,y)|x_iin X, y_jin Y }$, then i could always map $delta_{x_i}otimes delta_{y_j}$ into $delta_{x_i,y_j}(x,y)$ and get an isomorphism. $psi$ is one to one so it makes sense, but i can’t see how this fails if both X and Y have infinite elements.

I also tried other things like proving the surjectivity for $overline{phi}$, an element that can not be in the image of $psi$, or even showing $(overline{phi}, mathbb{R}^{Xtimes Y})$ didn’t have the universal property when both sets are infinite, but it didn’t seem to work for me.

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Any kind of help on this would be highly appreciated.