# Preimage of Subsets of $X_i$ in Infinite Cartesian Product

I ask this question only to have a clarification about the consequences of Halmos’ definition of Infinite Cartesian Product. Let $\{X_i\}_{i\in A}$ be an indexed family of subsets of a given set $X$. As you know $\displaystyle\prod_{\substack{i\in A}}X_i$ is defined to be the set of all functions with domain $A$ and codomain $\displaystyle\bigcup_{\substack{i\in A}}X_i$ such that $f(i)\in X_i ,\forall i\in A$. Now, let $\pi_i$ be the generic projection map and let $E\subseteq X_i$: how can $\pi_i^{-1}(E)$ be equal to $\displaystyle\prod_{\substack{j\in A}}E_j$ with $E_j=E$ if $j=i$ and $E_j=X_j$ otherwise? I mean, the codomain is relevant in this definition...

• I assume the $f$ is actually meant to be $\pi_i$. – Theo Bendit Sep 18 '18 at 10:12
• The result seems natural enough to me. What problem are you having with it? – Theo Bendit Sep 18 '18 at 10:16
• As the domain of f is A, $f^{-1}(E)$ is a subset of A. – William Elliot Sep 18 '18 at 10:20
• What is the final $f$ in $f^{-1}[E]$? – Henno Brandsma Sep 18 '18 at 10:51
• It was $\pi_i$ indeed, thanks for noticing! – Daniele Leo Sep 19 '18 at 5:32

I agree with the definition of the Cartesian product, that's completely standard (a set of functions $f$ from the index set $I$ such that $f(i) \in X_i$ for all $i \in I$). The codomain is "relevant" in the sense that it allows us to justify this as a set (from the axioms): a function is a subset of a product $A \times B$ (defined as sets of ordered pairs, e.g. with Kuratowski's definition $(a,b) := \{\{a\},\{a,b\}\}$ etc.) where $A$ and $B$ are already given to be sets. In this case we want $A = I$ (a set) and as $B$ we choose $\bigcup_i X_i$ (a set by the union axiom), because we already "know" that all values of $f$'s in the Cartesian product we want to define are going to be in this union. So this way we can garantuee from the pairs/union/comprehension etc. axioms that $\prod_i X_i$ is going to be a well-defined set. But "really" it's just some collection of ordered pairs fulfilling some conditions. I give these conditions at the end in more formal terms.

The projection $\pi_j: \prod_{i \in I} X_i \to X_j$, for any fixed $j \in I$ is defined as $\pi_j(f) = f(j)$, which indeed is in $X_j$ by definition.

If $E$ is a subset of $X_j$, then consider what it means to be some $f$ in $\pi_j^{-1}[E]$: $f \in \prod_{i \in I} X_i$ so it is a function defined on $I$ with $f(i) \in X_i$ for all $i$ and $\pi_j(f) \in E$ forces the only extra restriction that in fact $f(j) \in E$. But these are also exactly the conditions for a function on $I$ to be in $\prod_{i \in I} E_i$ where all $E_i = X_i$ except $E_j=E$. They're the same conditions on $f$ as a set of ordered pairs. Hence that we can say

$$\pi_j^{-1}[E] = \prod_{i \in I} E_i \text{ where all } E_i \text{ (except } E_j = E \text{) equal } X_i$$

So to recap a set $f$ is in $\prod_{i \in I} X_i$ iff

• $\forall x \in f: \exists i \in I: \exists p: x= (i,p)$.
• $\forall i \in I: \exists p: (i,p) \in f$.
• $\forall i \in I: \forall p,p': ((i,p) \in f \land (i,p') \in f) \implies p=p'$. (the last equality is equality of sets, so really $p=p'$ should be $\forall x: x \in p \leftrightarrow x \in p'$)
• $\forall i \in I: \exists x \in X_i: (i,x) \in f$.
• Functions in $\displaystyle\prod_{\substack{i\in A}}X_i$ have $\displaystyle\bigcup_{\substack{i\in A}}X_i$ as codomain; functions in $\displaystyle\prod_{\substack{i\in A}}E_i$ have $\displaystyle\bigcup_{\substack{i\in A}}E_i$ as codomain. How this can be formally contained in the first one? It’s just fussiness, I got all of the definitions, the concepts behind, just that missing point – Daniele Leo Sep 19 '18 at 5:48
• @DanieleLee A function is a set of pairs; its domain and codomain can be inferred from the set. – Henno Brandsma Sep 19 '18 at 11:41
• I got it! Thanks Henno! :) – Daniele Leo Sep 22 '18 at 10:43
• @DanieleLee glad to be able to help! – Henno Brandsma Sep 22 '18 at 13:02

It's ${\bf\pi_i}^{-1}(E)$ that is $\prod_jE_j$ with $E_j=X_j$ if $j\ne i$ and $E_i=E$.