Wikipedia:Reference desk/Archives/Mathematics/2016 February 19

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February 19

Cartesian product in constructive mathematics

Is there a constructive proof (i.e. a proof in constructive mathematics) of the fact that if a Cartesian product of sets is a singleton, then all of the sets are singletons? Classically, if $(x_{i})_{i\in I}$ is the unique element of the Cartesian product $\prod _{i\in I}X_{i}$, and $y\in X_{i}$, then one can consider the family $(y_{j})_{j\in I}$ where $y_{i}=y$ and $y_{j}=x_{j}$ if $j\neq i$, and from this deduce that $y=x_{i}$, showing that $X_{i}$ is a singleton for all $i\in I$. GeoffreyT2000 (talk) 23:25, 19 February 2016 (UTC)

I may be missing something stupid but it seems like your proof works constructively. Let the Cartesian product be $\{f\}$ where $f$ is the tuple as a function on $I$. I'll say $S$ is a singleton if $\exists x\,(S=\{x\})$, i.e. if $\exists x{\in }S\,\forall y{\in }S\,(x=y)$. Then you want to prove $\forall i{\in }I\,\exists x{\in }X_{i}\,\forall y{\in }X_{i}\,(x=y)$. The proof is: given $i$, take $x=f(i)$; given $y$, define $g(j)=(y{\mbox{ if }}j=i,f(j){\mbox{ otherwise}})$; then $g\in \{f\}$, so $g=f$, so $y=g(i)=f(i)=x$. -- BenRG (talk) 03:16, 22 February 2016 (UTC)