In "Lattices and Ordered Sets" author S. Roman defines the Cartesian product of a family of sets. I understand the concept. What I don't understand however is the notation he has used. He says, "for this, we use functions". Now see the definition on the attached copy of a page from his book starting with $ \Pi \mathcal{F}.$ Please explain in detail what all of the notation means, what is the domain, the range? ( I got many duplicate question candidates about Cartesian product families but none answered my question ).

page 1 Lattices and Ordered Sets

  • $\begingroup$ $\prod\mathcal{F}$ means exactly the same as the other two symbols. $\endgroup$ – egreg Jan 8 '15 at 16:34
  • $\begingroup$ I am basically interested what is in between the brackets. $\endgroup$ – nilo de roock Jan 8 '15 at 16:36
  • $\begingroup$ What isn't clear? $\endgroup$ – egreg Jan 8 '15 at 16:37
  • $\begingroup$ How it is a function, what maps to what? WHere does the f (i) come from. $\endgroup$ – nilo de roock Jan 8 '15 at 16:40
  • $\begingroup$ The $ I \mapsto $ is confusing. $\endgroup$ – nilo de roock Jan 8 '15 at 16:42

The notation $$ \biggl\{ f\colon I\to \bigcup_{i\in I}X_i\biggm| f(i)\in X_i\biggr\} $$ (there's an obvious typo, as $A_i$ should be $X_i$) denotes the set consisting of all functions having domain $I$ (the index set of the family $\mathcal{F}$) and codomain the union of the sets in the family $\mathcal{F}$ that satisfy the condition $$ f(i)\in X_i $$ for all $i\in I$ (the “for all” clause is also missing).

So, if $I=\{0,1\}$ and $X_0=\{a,b,c\}$, $X_1=\{c,d,e\}$, we should think to all functions $$ f\colon \{0,1\}\to\{a,b,c,d,e\} $$ such that $f(0)\in X_0$ and $f(1)\in X_1$. Thus $$ \prod_{i\in\{0,1\}}X_i $$ consists of the functions \begin{align} \Bigl\{& \{(0,a),(1,c)\}, \{(0,a),(1,d)\}, \{(0,a),(1,e)\},\\& \{(0,b),(1,c)\}, \{(0,b),(1,d)\}, \{(0,b),(1,e)\},\\& \{(0,c),(1,c)\}, \{(0,c),(1,d)\}, \{(0,c),(1,e)\}\Bigr\} \end{align}

  • $\begingroup$ Thank G__ a typo, you made it much clearer, thanks. $\endgroup$ – nilo de roock Jan 8 '15 at 16:58

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