The definition of the wiki is one that I find intutive, this is define the product space of $X_1,X_2,\dots$ by their cartesian product, this is, $X:=\prod_{a\in I} X_a$.

Now looking back to my notes, a book has given a definition of product space that I don't quite grasp, defining $X:=\prod_{a\in I}X_a = \{\omega:I\to\bigcup_{a\in I}X_a : \omega(a)\in X_a\;\forall a\in I\}$ where $I$ is the index set.

I have several questions here:

(1) Are we really talking about the same space?, or they just have the same name?. The first product set contain elements whose are elements of $X_a$, the second definitions looks like consider the elements of the set as functions and not 'regular' points.

(2) Also, it doesn't look very intutive as the picture of product set I have. Trying to visualize this definition I consider the plane $\mathbb{R}^2$, according to the former definition is $$\mathbb{R}^2=\prod_{i\in\{1,2\}}\mathbb{R_i} =\{\omega:\{1,2\}\to\mathbb{R_1}\cup\mathbb{R_2} : \omega(i)\in \mathbb{R}_i, i\in\{1,2\}\}$$

And I fail to see how this that looks like a space of functions will give the familiar $\mathbb{R}^2$.

  • 1
    $\begingroup$ You defined the second option. Can you do that too for the first (Cartesian product)? A look on math.stackexchange.com/a/771295/75923 might help. $\endgroup$
    – drhab
    Apr 30 '14 at 11:48
  • $\begingroup$ @drhab Thanks, that question looks exactly what I was looking for (and I couldn't find it using the search box!) so I'll give it a read. To answer your question, using the cartesian product I believe the definition would be given by $\mathbb{R}^2:=\{(a,b):a,b\in \mathbb{R}\}$. This would work?, because this is how I think intuitively of any product space (this is, the n-tuples all possible combinations of elements of the set such that the $i$ entry of the n-tuple is in the $X_i$ space). $\endgroup$
    – Cure
    Apr 30 '14 at 12:04
  • $\begingroup$ Your intuitive thinking is okay. But troubles arise when the indexset becomes infinite (see the answer of Ittay). When it comes to $n$-tuples then here is another link that might interest you: math.stackexchange.com/q/721401/75923 $\endgroup$
    – drhab
    Apr 30 '14 at 12:19

Firstly, this is an issue of elementary set theory, not topology, since you are just wondering about the definition of the caretesian product of sets.

As long as you have finitely many sets, there is little doubt as how to define, say, $A\times B$. Now, the naive definition is that it is the set $\{(a,b)\mid a\in A,b\in B\}$. This is not a formal definition though since we need to be precise about what $(a,b)$ means. There are several ways around that, one is to define $(a,b)=\{\{a,b\}, \{a\}\}$. It is an easy exercise to see that this indeed behaves like we would expect. Then you the definition of the cartesian product above becomes precise, and it can easily be adapted to any finite number of sets.

Now, things get more complicated already when we consider countably many sets. One usually says that the product $A_1 \times \cdots A_n \times \cdots $ is the set whose elements are $(a_1, \ldots, a_n,\ldots)$, with $a_k\in A_k$ for all $k\ge 1$. However, to make this into a precise definition, what exactly is this long list of elements? Quite often one says "I don't care!" and just proceeds to work with such longs lists, without worrying about foundational issues. There is nothing wrong with such an attitude. However, when considering the cartesian product of more than countably many sets, one really can't just say "well, we'll take really long sequences of elements this time".

Fortunately, there is a way to work with cartesian products of any family of sets, and that is the definition by functions you gave. It works uniformly for all families of sets. Now, $\mathbb R^2$ according to that definition consists of all functions $f:\{1,2\}\to \mathbb R$. Such a function is uniquely determined by (ta da ta da ta daaaaaa) a pair of real numbers. More formally, given such an $f$, consider the pair $(f(1),f(2))$. Consider a pair or real numbers $(a,b)$, then you get the corresponding function $f(1)=a$ and $f(2)=b$. So, it essentially is just the good old $\mathbb R^2$ you know and love.


For this occasion let $\mathbb{R}^{2}$ denote here the space defined under (2) in your question and let $\mathbb{R}\times\mathbb{R}=\left\{ \left(x,y\right)\mid x\in\mathbb{R},y\in\mathbb{R}\right\} $ denote the 'normal' cartesian product.

These spaces do not have the same elements. Both can serve as product for pair $\left(\mathbb{R},\mathbb{R}\right)$.

The first one with projections $\mathbb{R}^{2}\rightarrow\mathbb{R}$ prescribed by $\omega\mapsto\omega\left(i\right)$ for $i=1,2$ and the second with projections $\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ prescribed by $\left(x_{1},x_{2}\right)\mapsto x_{i}$.

Function $\phi:\mathbb{R}^{2}\rightarrow \mathbb{R}\times\mathbb{R}$ defined by $\omega\mapsto\left\{ \omega\left(1\right),\omega\left(2\right)\right\} $ is a bijection. So the spaces are isomorphic.

Spaces defined as $\mathbb{R}^{2}$ become 'fruitful' if you are dealing with infinite indexsets.


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