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I was reading the Princeton Review of GRE math subject test (4th edition), and one question was (page. 251)

Example 6.24 Is the ring $\mathbf{R}^\mathbf{R}$ an integral domain?

Solution: Consider the function $f:\mathbf{R} \rightarrow \mathbf{R}$ and $g:\mathbf{R} \rightarrow \mathbf{R}$ given by $$f(x) = x - |x| \quad \text{and} \quad g(x) = x + |x|$$ The pointwise product function, $fg$, is the zero function--that is, $(fg)(x)=0$ for every $x$ in $\mathbf{R}$--even though neither $f$ nor $g$ is the zero function. Since $\mathbf{R}^\mathbf{R}$ contains zero divisors, it's not an integral domain.

I did not find this notation anywhere else in this book, and I tried to guess, but cannot come up with an idea. Please help, thanks in advance.

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  • $\begingroup$ $X^Y$ is the set of all functions from $Y$ to $X$. $\endgroup$ – Thomas Andrews Apr 14 '15 at 23:12
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For general sets, $A^B$ denotes the set of functions from $B$ into $A$. In your case, $\Bbb R^{\Bbb R}$ denotes the set of functions from $\Bbb R$ to $\Bbb R$.

A way to see that this makes sense is that $\Bbb R^3$ is usually viewed as a vector space with 3 basis vectors. We write a general element as $(x_1,x_2,x_3)$. We could instead view this as a function from the three element set to $\Bbb R$ by $f(1) = x_1$, $f(2) = x_2$ and $f(3) = x_3$. This realization of functions (or vectors, etc) allows for some things in topology to be stated quite simply.

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  • $\begingroup$ Thank you! I really like your $\Bbb R^3$ example. Actually at first I thought about $\Bbb R$ dimensional $\Bbb R$ space but I did not realize what it is. $\endgroup$ – MonkeyKing Apr 14 '15 at 23:27
  • $\begingroup$ It's worth mentioning that $|X^Y| = |X|^{|Y|}$. $\endgroup$ – GFauxPas Apr 15 '15 at 0:30
  • $\begingroup$ and even further currying: $|(X^{Y})^{Z}|=|X|^{|Y||Z|}$ $\endgroup$ – Ali Caglayan Mar 2 '17 at 17:48

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