Is choice required to guarantee that $\mathbb R^2 := \mathbb R \times \mathbb R$ – or $\mathbb R^n := \displaystyle\prod_{k=1}^n \mathbb R$ in general – isn't the empty set $\varnothing$? If not, what's an example of a Cartesian product of two non-empty sets not guaranteed to be non-empty (without choice)?
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You do not need the axiom of choice to take Cartesian products of finite sets. More on this at https://en.wikipedia.org/wiki/Axiom_of_choice – parsiad Apr 08 '18 at 21:29
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$(0,\dots,0)$ is always an element of $\mathbb R^n$. – Wojowu Apr 08 '18 at 21:31
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You don’t need the axiom of choice to prove thta an arbitrary power of a single nonempty set is not empty. – Arturo Magidin Apr 08 '18 at 21:32
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1@parsiad But $\mathbb R$ is not finite. – Markus Klyver Apr 08 '18 at 21:33
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@MarkusKlyver: A product of finitely many nonempty sets is nonempty; you do not need AC to establish this, even if the sets themselves are infinite. For some sets, you do not need AC at all even when there are infinitely many factors, but then it depends on the factors. – Arturo Magidin Apr 08 '18 at 21:35
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1@MarkusKlyver: you're right; I meant finitely many. I guess this got answered below anyways. – parsiad Apr 08 '18 at 23:45
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If $X$ is nonempty and $\kappa$ is an arbitrary cardinal, then you do not need $AC$ to prove that $X^{\kappa}$ (the cartesian power of $X$) is nonempty: since $X$ is nonempty, there exists $x_0\in X$. Then define $f\colon\kappa\to X$ by $f(i) = x_0$ for all $i\in\kappa$. This is an element of $X^{\kappa}$, so the latter is not empty. No choice needed.
(If $\kappa=0$, you get the empty function, which is the unique element of $X^{\varnothing}$)
Arturo Magidin
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The ability to make (and remember) finitely many choices during a proof is built into the rules of first-order logic. You don't need a specific set-theoretic axiom to do that.
By induction, you can then prove in general that choice functions for a family of sets you know has finite cardinality must also exist.
Independently of this, $A^B$ is always nonempty when $A$ is. This also does not require any axiom of choice -- you can just choose one element $a\in A$ and then consider the function that maps every element of $B$ to this $a$.
hmakholm left over Monica
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@MarkusKlyver: You may use choice for infinite products, but nobody says you always have to. In some cases you know enough about the factors that you can get a different proof without choice to work -- such as here where you know that the intersection of all the factors is non-empty. – hmakholm left over Monica Apr 08 '18 at 21:41
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So without choice, the claim that a particular product is non-empty could actually be independent of the axioms (and hence you would need choice)? – Markus Klyver Apr 08 '18 at 21:47
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1x @Markus: Yes -- for example this is the case for $ \prod\limits_{i\in I} i$ where $I=\mathcal P(\mathbb R)\setminus{0}$. – hmakholm left over Monica Apr 08 '18 at 21:51
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I loved your use of the word "remember": I think that this is the gist of the question. – Matemáticos Chibchas Mar 21 '19 at 17:58
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Choice is not required for finite products of sets.
In the particular case of $\mathbb{R}^2$ it's even easier: $(0,0)\in\mathbb{R}^2$.
egreg
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2@MarkusKlyver: No: cartesian powers of nonempty sets are nonempty; no need of AC to establish that. – Arturo Magidin Apr 08 '18 at 21:34
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@MarkusKlyver Constant functions don't need choice. More generally, if you have a family of sets $(A_\lambda){\lambda\in\Lambda}$ so that $\bigcap{\lambda\in\Lambda} A_\lambda$ is nonempty, then $\prod_{\lambda\in\Lambda} A_\lambda$ is nonempty: just take an element $x$ in the intersection and define the function $f\colon\Lambda\to\bigcup_{\lambda\in\Lambda}A_\lambda$ such that $f(\lambda)=x$, for every $\lambda\in\Lambda$. – egreg Apr 08 '18 at 21:46