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

= 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)