Indexed System of Sets Proof Explanation

Question

Proof:

1. I'm not sure what $f_i$ is. is $f_i$ an ordered pair or an element of the codomain of f or what else?

2. Since f is a function on I, is I the codomain or domain of function f?

3. What is I?

   4. What does the exponentiation or multiplication of sets yield? Is the exponentiation or multiplication of sets the same thing as doing the infinite or n-ary cartesian product?

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Ignore the background below if you know what we're talking about:

Answer 1

Long story short: $f\subset I\times \cup_{i \in I}S_i$ means that $I$ is domain and $\cup_{i \in I}S_i$ is codomain. $f_i\in S_i$, so $f_i$ is an element of codomain. Moreover, $f(i)=f_i$ from definition. It is true, that $(i, f_i) \in f$, but in proof there is different notation, namely $(i, s_i) \in f$. I admit this notation is somewhat confusing.

It is helpful to start from beginning and carefully build things up. First of all $I$ is an arbitrary nonempty set. Then there is some auxiliary nonempty set of sets (aka family of sets) call it $X$. Now we need to fix some function $\Phi$ with domain $I$ and codomain $X$, thus $\Phi:I\to X$. In set language $\Phi=\{(i,x):i\in I ,x\in X \}$. Looks familiar? Observe, that if we take $X:=\{ S_i:i\in I\}$ and define $\Phi(i):=S_i$, then $\Phi=S$. It can be helpful to look at the case when $I=\{1,..,n\}$, and $|X|=m$ is also finite. Fix $\Phi: \{1,..,n\}\to X$, if $\Phi$ is surjection (so $n\geq m$), all elements of $X$ can be listed (probably many times) $X=\{\Phi(1),..,\Phi(n)\}$. Now take $X_i:=\Phi(i)$. For finitary product there is no need to introduce all that stuff, just simple induction:

• $X_1\times X_2=\widehat\prod_{i=1}^2X_i=\{(x_1,x_2):x_1\in X_1,x_2\in X_2\}$

• $\widehat\prod_{i=1}^nX_i=(\widehat\prod_{i=1}^{n-1}X_i)\times X_n$

(There should be big "$\times$" sign instead of $\widehat\prod$, like $\textbf{here}$.) What is important, in finitary case induction based definition is equivalent to the "functional" one (there is cannonical bijection):

• $\Pi_{i=1}^nX_i=\{f:f\ is\ a\ function\ on\ \{1..n\}\ and\ f(i)\in X_i\}$

It is helpful to see what is inside though:

$$\widehat\prod_{i=1}^nX_i=\{(x_1,..,x_n):x_i\in X_i\}$$ $$\prod_{i=1}^nX_i=\{\{(1,x_1),..,(n,x_n)\}:x_i\in X_i\}$$

If induction based product is so simple, couldn't it be extended to arbitrary indexing set? It cannot and $\underline{\textbf{here}}$ is why.

Exponent analogy to numbers (or more general cardinals) is that $|\{f:\ f\subset A\times B ,\ f\ is \ function\}|=|B|^{|A|}$, see $\underline{\textbf{here}}$.