Testwiki: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 $j\ne i$, 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 $\mathrm{\exists }x(S=\{x\})$, i.e. if $\mathrm{\exists }x\in S\mathrm{\forall }y\in S(x=y)$. Then you want to prove $\mathrm{\forall }i\in I\mathrm{\exists }x\in X_i\mathrm{\forall }y\in X_i(x=y)$. The proof is: given $i$, take $x=f(i)$; given $y$, define $g(j)=(y\text{ if }j=i,f(j)\text{ otherwise})$; then $g\in \{f\}$, so $g=f$, so $y=g(i)=f(i)=x$. -- BenRG (talk) 03:16, 22 February 2016 (UTC)