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Suppose $\{ X_\alpha \}_{\alpha \in A} $ is a family of sets indexed by $A$. If $A $ is $\mathbb{N}$, then the Cartesian product of them is just

$$ X_1 \times X_2 \times \cdots $$

Reading Folland's book, he defines the Cartesian product of them $\prod_{\alpha \in A} X_{\alpha } $ as

$$ \{ f: A \to \bigcup_{\alpha \in A} X_\alpha : f( \alpha) \in X_\alpha \; \; \forall \alpha \} $$

I am having a hard time understanding this definition. Can someone explain it to me? I was trying to relate this definition to the case when $A = \mathbb{N}$ but still I am somewhat lost.

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This is really just the correct generalization of ordered pairs.

We think about an element of this product as some infinite tuple, and that's fine. But how do you define that?

We have a notion for ordered pairs, then we often define $A\times B\times C$ as $(A\times B)\times C$ or as $A\times (B\times C)$, and then we show that there is a natural bijection between those so it ultimately doesn't matter how we chose to define triplets.

Then by induction we extend this notion to every finite $n$. But just because we can do something by induction for every finite number doesn't mean we can do it for an infinite number of things at once. What would be an $\Bbb N$-tuple? Just an infinite list of open parentheses, then an element $x_1$, an infinite list of parentheses, $x_2$, and so on? This is problematic from several points of view. Including the modern set theoretic one, which defines $(a,b)=\{\{a\},\{a,b\}\}$ (this is how we interpret ordered pairs as sets), and defining an infinite tuple like that would cause an infinite sequence of sets $A_{n+1}\in A_n$, which contradicts the axiom of regularity.

It is much much simpler to observe that a $2$-tuple is just a function from an index set into some sets. So the first coordinate is always chosen from $A$ and the second from $B$; then we can easily extend this notion to larger domains, $A\times B\times C$ would be the set of functions from $\{1,2,3\}$ such that $f(1)\in A$, $f(2)\in B$ and $f(3)\in C$. And so on. So we can write it as functions from $\{1,2,3\}$ into $A\cup B\cup C$ such that $f(0)\in A$, $f(1)\in B$ and $f(2)\in C$.

And this notion easily extends to infinite products as well. So now we can define $\prod_{n\in\Bbb N}A_n$ as the functions from $\Bbb N$ into $\bigcup_{n\in\Bbb N}A_n$ such that $f(n)\in A_n$. And that's exactly what's there.

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Think of the Cartesian product of 3 sets. It's the set of ordered triples, $(x_1,x_2,x_3)$, where $x_i\in X_i$ for $i=1,2,3$. That means you can also describe it as the set of functions from $\{1,2,3\}\to \bigcup X_i$, with the restriction that $f(1)\in X_1$, $f(2)\in X_2$ and $f(3)\in X_3$. Just think of the ordered triple $(a,b,c)$ as the function given by $f(1)=a$, $f(2)=b$, $f(3)=c$.

Now let the set $A$ be much larger than just $\{1,2,3\}$. The definition generalizes perfectly.

Does that help?

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An element of $\prod_{\mathbb N} X_i$ is a list $(x_1,x_2,\ldots)$. What's such a list? It's a choice, for every $i\in \mathbb{N}$, of an element $x_i\in X_i$. What's a function from $\mathbb{N}$ to $\bigcup X_i$ such that $f(i)\in X_i$? It's a choice, for every $i\in\mathbb{N}$, of an element $x_i\in \bigcup X_i$ such that $x_i\in X_i$.

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$$ \left\{ f: A \to \bigcup_{\alpha \in A} X_\alpha : f( \alpha) \in X_\alpha \; \; \forall \alpha \right\} $$

In the exression $X_1\times X_2\times X_3\times\cdots$, the set $A$ is $\{1,2,3,4,\ldots\}$, and the function $f$ corresponding to $(x_1,x_2,x_3,\ldots)$ is given by $f(1)=x_1,\ f(2)=x_2,\ f(3)=x_3,\ \ldots\ {}$.

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  • $\begingroup$ Hi. This makes sense. Thanks for your reply. If you don't mind me asking, how can we think of this definition if $A$ is a set different than the natural numbers. For instance, say $A$ is the set of all continuous functions. How to make sense of this? $\endgroup$ – ILoveMath Oct 30 '14 at 4:01
  • $\begingroup$ You would have to assign some set $X_\alpha$ to every continuous function $\alpha$. Then a member of the Cartesian product would assign some member $x_\alpha$ to every continuous function. $\endgroup$ – Michael Hardy Oct 30 '14 at 4:07
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Say you take all the sets $X_\alpha$ and toss their contents into a hat. That's the union of the $X_\alpha$'s. Consider a function $f$ that, for each index $\alpha \in A$, picks some element out of the hat. Clearly there will be many such functions.

For a given function $f$ and a given index $\alpha$, since $f(\alpha)$ is in the union of all the sets $X_\alpha$, it may or may not be the case that $f(\alpha)$ happens to be an element of $X_\alpha$.

If a function $f$ has the property that for every $\alpha$, $f(\alpha) \in X_\alpha$, then we call $f$ a choice function. That's because we can think of $f$ as choosing an element out of each of the sets $X_\alpha$.

If you work out this example with a finite or countable collection of $A_\alpha$'s, it corresponds to the usual intuition of the Cartesian product as a set of n-tuples or even countably infinite-tuples.

The choice function definition simply extends this idea to arbitrary index sets. In fact if you think about a choice function as an "A-tuple" it's exactly the same.

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