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$\begingroup$

$\mathbb{R}^\mathbb{R}$ being the ring of all functions $\mathbb{R}\to\mathbb{R}$ with multiplication and addition defined as follows:

(f+g)(x) = f(x) + g(x)

(fg)(x) = f(x)g(x)

I know that $\mathbb{R}^\mathbb{X}$ is a ring when $\mathbb{X}$ is a finite set, but I'm not quite sure how the axioms would work with infinite sets.

Thank you

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    You can think of $\mathbb{R}^\mathbb{R}$ as the direct product $\prod_{i\in\mathbb{R}}\mathbb{R}$, if that helps. But yes, it is a ring. – 2017-01-08
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    Is there a ring axiom that fails? – 2017-01-08
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    In fact, for any $Y$, $X^Y$ is a ring whenever $X$ is a ring. – 2017-01-08
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    Sum and multiplication of real numbers are reals numbers too, so... – 2017-01-08
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    @OpenBall ... Contrariwise, I would say $X^Y$ is a ring whenever $X$ is a ring. – 2017-01-08
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    @GEdgar I always forget which one is the domain and which one is the codomain in $X^Y$.. – 2017-01-08
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    @OpenBall I was just about to correct that too. Also, *Y* needs to be a nonemplty set. Forgot to write that I knew that. Thanks though! – 2017-01-08
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    @LeonSot got it, thanks – 2017-01-08

2 Answers 2

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Yes, it is a ring.

Since $(\mathbb R^{\mathbb R},+)$ is an abelian group (the inverse of a function $f$ is...$-f=[x\mapsto -f(x)]$). And the other axioms are verified too.

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More general if $\mathcal{F}(X,\mathbb{R})$ with $X\ne \emptyset $ is the set of all functions from $X$ to $\mathbb{R}$ then, $\left(\mathcal{F}(X,\mathbb{R}\right),+,\cdot)$ is an unitary and communtative ring.

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    Strange for you to switch notation from that used by the question – 2017-01-08