# confusion regarding infinite cartesian product

Suppose I am given an infinite collection of $$R$$-modules $$\{M_i\}_{i\in I}$$. Their infinite cartesian product is $$\prod_{i\in I} M_i$$ and also can have an $$R$$-module structure. If I am given a bijection $$\phi:I \to I$$, then there is an $$R$$-isomorphism between $$\prod_{i\in I} M_i$$ and $$\prod_{i\in I} M_{\phi(i)}$$.

My confusion here is that in the finite case, where we have some order to the elements, there is something to show. But here in the infinite case it seems too obvious, since I don't have any particular order in $$I$$ that I follow. An element in $$\prod_{i\in I} M_i$$ is just some $$\{m_i\}_{i\in I}$$. So why would changing the "order" should make any difference? (Note: This question may be also relevant to an infinite cartesian product of sets, but I stumble upon it in context of modules, so I left the question as it is)

• @SassatelliGiulio this is exactly what I mean. The isomorphism doesn't do anything to the function, everything stays the same Nov 17, 2023 at 21:35
• Here's the secret: even in the finite case there isn't an order. Nov 17, 2023 at 21:45
• @JonathanZ how do you define the finite case? Similarly to the infinite? Nov 17, 2023 at 21:50
• @User666x - (Sorry for just dropping that cryptic comment. I wanted you to stick around while I wrote up my answer. :-) ) Nov 17, 2023 at 22:02
• @User666x $\{m_i\}_{i\in I}$ is actually abuse of notion. Elements of cartesian product are function with certain condition. Jan 12 at 14:10

Given a function $$N_\bullet:I\to \text{Modules}$$ (but really to Sets), $$\prod_{i\in I} N_i$$ is the set of functions $$f:I\to\bigcup_{i\in I} N_i$$ such that $$\forall i\in I, f(i)\in N_i$$. So if you specialise to the case $$N_\bullet=M_{\phi(\bullet)}$$ you obtain that $$\prod_{i\in I} M_{\phi(i)}$$ is the set of functions $$f:I\to \bigcup_{i\in I}M_{\phi(i)}$$ (which is indeed equal to $$\bigcup_{i\in I}M_{i}$$ in this case) such that $$\forall i\in I, f(i)\in M_{\phi(i)}$$. As opposed to $$f(i)\in M_i$$.

Now you can see some point to having a map $$\phi^*:\prod_{i\in I} M_i\to \prod_{i\in I} M_{\phi(i)}$$, $$\phi^*(f)=f\circ\phi$$. Or, more similar to your notation, $$\{(\phi^*m)_i\}_{i\in I}=\{m_{\phi(i)}\}_{i\in I}$$. Have fun proving that it's well-defined and bijective.

• Excellent, thank you! I understand why I got it wrong Nov 17, 2023 at 21:42

Here's the secret: even in the finite case there isn't an order.

Let's take the case of $$\mathbb R \times \mathbb R \times \mathbb R$$. It's actually the set of all maps from $$\{1,2,3\}$$ to $$\mathbb R$$. Let's look at one of them: call it $$f$$, and $$f$$ maps

$$1 \mapsto e\\ 2 \mapsto \pi\\ 3 \mapsto 17$$

Well, that's a pain to write. So we decide to use the convenient fact that there is a standard ordering on $$\{1,2,3\}$$ that everyone agrees on, and shorthand $$f$$ as $$(e, \pi, 17)$$.

But notice two things:

• $$f$$ would still be the same thing if we described it as

$$2 \mapsto \pi\\ 3 \mapsto 17\\ 1 \mapsto e\\$$

• And, if we had decided to treat $$\mathbb R \times \mathbb R \times \mathbb R$$ as the set of all maps from $$\{cat, elevator, cheese\}$$ to $$\mathbb R$$, we'd have to stop using the ordered pair notation.

Basically, my claim (and even I realize that it comes close to tin foil hat-level paranoia) is that because we write text sequentially from left to right, we easily slip into using that order for our notation, and we think that the order is essential to our math, but it's actually just a notational convenience. It's only when we start looking at infinite sets that we can't explicitly write down (like you just did) that we can notice that that's what we did.

(Bonus secret: All that linear algebra stuff, where we're writing up the matrix of a transformation w.r.t. a basis? It's actually w.r.t. an ordered basis. A basis is defined as an (unordered) set, but we just write "$$v_1, v_2, v_3, ...$$, and an ordering slips in there too!)

• You are absolutely right! Nov 17, 2023 at 22:05