The definition of the product of sets

Question

I got a definition :

If I is a nonempty set and for each i in I we have a nonempty sets $ X_i$, then the product of these sets is defined by $$\prod_{i\in I}X_i=\{x:I \rightarrow \bigcup_{i\in I} X_i:x(i)\in X_i\text{ for all } i\ \in I\}$$

The definition is from [https://www.springer.com/gp/book/9783319023670 ]. I think this is not easy to understand it.

Assume we want to prove :

Suppose I is a set, $X_i$ are topological spaces with subsets $A_i$ for $i \in I$. Then, there holds $\operatorname{Clos}\prod_{i\in I}A_i=\prod_{i\in I}\operatorname{Clos} A_i$

Then, the proof $$\prod_{i\in I}\operatorname{Clos} A_i$$ $$=\{x:I \rightarrow \bigcup_{i\in I} \operatorname{Clos} A_i : x(i)\in \operatorname{Clos} A_i\ \text{ for all } i \in I\} $$ $$=\{x:I \rightarrow \operatorname{Clos}(\bigcup_{i\in I} A_i) :x(i)\in \operatorname{Clos} A_i \text{ for all } i \in\ I\}$$ $$=\operatorname{Clos} \prod_{i\in I}A_i$$ is right?

Answer 1

No, closure does not commute with infinite unions as you want to use in the step from line 2 to line 3. The last step is also nonsense, as the closure of a product set depends on the topology we put on the product: it's not a pure set theory fact.

You need to show two inclusions and use the definition of closure and the specific form of open sets in the product topology.

If you have covered nets (generalised sequences) in your course, the following proof does work (modulo some minor details, but in essence):

$x \in \operatorname{Clos} \prod_i A_i$ iff there is some net $(a_n)_{n \in N}$ where all $a_n \in \prod_{i \in I} A_i$ and $a_n \to x$ iff for all $i \in I$ $(a_n)(i) \to x(i)$ iff for all $i \in I$, $x(i) \in \operatorname{Clos} A_i$ iff $x \in \prod_i \operatorname{Clos} A_i$.

Answer 2

Your solution is making the assumptions that

1) $\bigcup\limits_{i\in I}ClosA_i=Clos\bigcup\limits_{i\in I}A_i$ but you can easily see that what holds, in general, is that $\bigcup\limits_{i\in I}ClosA_i\subseteq Clos\bigcup\limits_{i\in I}A_i$ where the equality might not hold in general, eg consider $A_i=(\frac{1}{i+1},\frac{1}{i}),i\in \mathbb{N}$ then, $\bigcup\limits_{i\in \mathbb{N}}ClosA_i=(0,1]\neq Clos\bigcup\limits_{i\in \mathbb{N}}A_i=[0,1]$.

2)$Clos\prod\limits_{i\in I}A_i=Clos\{x:I\rightarrow \bigcup\limits_{i\in \mathbb{N}}A_i|x(i)\in A_i \forall i\in I\}=\{x:I\rightarrow Clos\bigcup\limits_{i\in \mathbb{N}}A_i|x(i)\in ClosA_i \forall i\in I\}$ which does not make much sense as now you must consider closure of a set of functions but you must know the topology on them to know how to deal with them.

I know of a way to prove the aforementioned result in the product or box topology using the fact that $x\in ClosA\ iff\ for\ any\ U\ni x\ open\ in\ X, U\cap X\neq \phi$, you can find the same in munkres at the end of the section on product topology(section-19). Though I'm pretty sure it should be provable this way also, I'm not able to come up with analogs for the product or box topology in this functional form, i.e., you have to show that $$Clos\{x:I\rightarrow \bigcup\limits_{i\in \mathbb{N}}A_i|x(i)\in A_i \forall i\in I\}=\{x:I\rightarrow \bigcup\limits_{i\in \mathbb{N}}ClosA_i|x(i)\in ClosA_i \forall i\in I\}$$