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I first learned about the Cartesian product as ordered pairs, but soon when we need to get the infinite Cartesian product, we regard it as mapping. I.e.$\prod U_i(i \in I)$= $(f:I \longrightarrow \cup_{i \ in I} U_i \wedge f(i)\in U_{i})$. I know that in essence, they are talking about the same thing, but how can a mapping be equivalent to ordered pairs? It is quite confusing.

My question is

Why a set of mappings is equivalent to a set of ordered pairs?

Definition of the Infinite Cartesian Product fully solved my question.

Andrew_Ren
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1 Answers1

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The ordered pair $(x,y)\in E^2$ can be identified to the mapping $\{0,1\}\to E,\;0\mapsto x,\;1\mapsto y.$

Anne Bauval
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  • In computer science, this is similar to having a pair of pointers to 2 objects, instead of having a pair of objects. – Jean-Armand Moroni Sep 14 '22 at 14:06
  • Yes, I understand. But when proving for example $(A\cap B) x (C \cap D)=(A x C) \cap (B x D)$, the ‘mapping’ definition seems unreachable – Andrew_Ren Sep 14 '22 at 14:06
  • $$\left[(f(0)\in A\cap B)\land(f(1)\in C\cap D)\right]\Leftrightarrow\left[\left(f(0)\in A\land (f(1)\in C)\right)\land\left(f(0)\in B\land (f(1)\in D)\right)\right].$$ – Anne Bauval Sep 14 '22 at 14:18