Cartesian Product Definitions

[Calculuser]

I was studying Group Theory on my own from a mathematics journal and got confused at some point where it defines Cartesian products, from binary one, say ($A × B$), to n-tuples one, say ($A_1 × A_2 × ... × A_n$). What confuses me when I tried to read it is that the definition made for infinite Cartesian product as shown below:

If I am correct that on the left-hand side $\prod_{i\in{I}}X_i$ by definition of Cartesian product corresponds to $X_1 × X_2 × ...$, which can be represented as in $(x_1,x_2,...)$. However, when I try to read and interpret the right-hand side, I fail to create those tuples $(x_1,x_2,...)$, but think of a set of $\{f_i\}$, which seemed to me absurd that how this set can turn into tuples.

I would like to know how to interpret all.

 

[fresh_42]

I was studying Group Theory on my own from a mathematics journal and got confused at some point where it defines Cartesian products, from binary one, say ($A × B$), to n-tuples one, say ($A_1 × A_2 × ... × A_n$). What confuses me when I tried to read it is that the definition made for infinite Cartesian product as shown below:
View attachment 233130
If I am correct that on the left-hand side $\prod_{i\in{I}}X_i$ by definition of Cartesian product corresponds to $X_1 × X_2 × ...$, which can be represented as in $(x_1,x_2,...)$. However, when I try to read and interpret the right-hand side, I fail to create those tuples $(x_1,x_2,...)$, but think of a set of $\{f_i\}$, which seemed to me absurd that how this set can turn into tuples.

I would like to know how to interpret all.

To interpret it correctly, I short detour in category theory might be necessary. You can find the correct definition of a direct product and a direct sum, which is something different, on Wikipedia. I quote Wikipedia here, as it was even linked to on nLab, which is in general the better source for mathematical definitions.

Now to your question. The right hand side of your formula, unfortunately the one you have trouble with, is the correct one. $n-$tuples are o.k. if $n$ is a finite number, but for $n=\infty$ they are only second best, because they suggest an ordering and a countability which the index set in the direct product doesn't need to have. That's why the elements of a direct product are represented by functions instead, as they don't need this list: $1,2,3,4,5,6,7,\ldots$

So let $f \in \Pi_{\iota \in I}X_\iota$, a function which assigns an $x_\iota = f(\iota) \in X_\iota$ to every possible index $\iota$. You see, that in case $I=\{\,1,2,\ldots ,n\,\}$ we can write $f$ as the list of its images: $\{\,f(1),f(2),\ldots,f(n)\,\}=\{\,x_1,x_2,\,\ldots ,x_n\}$ and in case of countable infinity even by replacing $f(n)=x_n$ by dots. But what if $I=\mathbb{R}\,?$ Then we cannot write down the images of all $f(r)$, so our element in the direct product has to be written as this function $f$, a certain selection of an $x_r \in X_r$ for all indices $r \in I=\mathbb{R}$, for short $f=(f(r))_{r\in \mathbb{R}}$.

 

[Calculuser]

First of all, I would like to thank you in advance for the reply before my further inquiry. I have a question in mind regarding my way of thinking that needs to be confirmed or corrected if I am wrong.

What is the role of $(\forall{i}))(f(i)\in{X_i})$ statement in the set in question?

I removed that part from the set definition in question and then pondered what it would mean only with the rest. It would mean nothing but a list of functions defined as $I → \bigcup_{i\in{I}}X_i$. Therefore, if there existed a function, say $f_1(i)$ with $I = \{a,b\}$ and $X_a$, $X_b$, such that it mapped all elements to arbitrary element(s) under the condition of $f_1(a), f_1(b) \in X_a$, which in my opinion makes the existence of $X_b$ immaterial. That is why I thought we need to modify that definition by adding the statement in the question above that necessitates the selection of an element from $X_b$. In other words, I thought of functions as choice functions that picks out elemets from each $X_i$.

Is there any missing point above?

Additionally, If we write the expression as $\prod_{i\in{I}}X_i=(x_1,x_2,...)=\{f_1,f_3,f_4,...\}$, then how can we say that ordered tuples defined by $(...)$ are equivalent to a set defined by $\{...\}$? They are different notations. Do we just ignore that?

 

[fresh_42]

First of all, I would like to thank you in advance for the reply before my further inquiry. I have a question in mind regarding my way of thinking that needs to be confirmed or corrected if I am wrong.

What is the role of $(\forall{i}))(f(i)\in{X_i})$ statement in the set in question?

It specifies the element. In general, a notation $M:=\{\,a \in S\,|\,P(a) \text{ true }\,\}$ reads: "The set of all elements $a$ from $S$, which additionally satisfy property $P(a)$. The part $\in S$ can be omitted, if it is clear which set we are talking about, or substituted by a description of $S$ as in our example. Also the part $\text{ true }$ is normally omitted, as it is implied by the location within the definition in $\,|\,\ldots \,\}$. Here are all necessary conditions listed, which make the $a$ elements of $M$ instead of just elements of $S$. In our case we have:##

• $M \triangleq \Pi_{\iota \in I}X_\iota$
• $a \triangleq f$
• $S \triangleq \text{The set of all functions }I \longrightarrow \bigcup_{\iota \in I}X_\iota$
• $P(a) \triangleq f\text{ can only assign an element from the set }X_\iota \text{ for every }\iota \in I \Longleftrightarrow (\forall{\iota})(f(\iota)\in{X_\iota})$

I removed that part from the set definition in question and then pondered what it would mean only with the rest. It would mean nothing but a list of functions defined as $I → \bigcup_{i\in{I}}X_i$. Therefore, if there existed a function, say $f_1(i)$ with $I = \{a,b\}$ and $X_a$, $X_b$, such that it mapped all elements to arbitrary element(s) under the condition of $f_1(a), f_1(b) \in X_a$, which in my opinion makes the existence of $X_b$ immaterial. That is why I thought we need to modify that definition by adding the statement in the question above that necessitates the selection of an element from $X_b$.

?

In other words, I thought of functions as choice functions that picks out elemets from each $X_i$.

Yes. And you cannot pick from $X_j$ at the $i-$th position if $i\neq j\,.$

E.g. If we have $X_1=\{\,a,b\,\}\, , \,X_2=\{\,\text{ trunk },\text{ branch },\text{ leaf }\,\}$, then $f(1)=b\, , \,f(2)=\text{ trunk }$ is allowed, and since we have finitely many indices, i.e. $|I|=2\,,$ we may write $f \triangleq (b,\text{trunk})$. Elements $g(1)=b\, , \,g(2)=a$ or $h(1)=\text{leaf}\, , \,h(2)=\text{trunk}$ are not allowed in the direct product. If we removed the $P(a) \text{ true }$ part of the definition, as you suggested, the functions $g$ and $h$ would be allowed. Thus the part ##P(a) \triangleq

Is there any missing point above?

I don't know, because I did not understand what you wrote. I hope I have answered it anyway.

Additionally, If we write the expression as $\prod_{i\in{I}}X_i=(x_1,x_2,...)=\{f_1,f_3,f_4,...\}$, then how can we say that ordered tuples defined by $(...)$ are equivalent to a set defined by $\{...\}$? They are different notations. Do we just ignore that?

First of all, this will only work for countable index sets $I$, not for index sets like $I=\mathbb{R}$. For the arguments sake, let us assume $I=\mathbb{N}$ here! You are right, that $(x_1=f(1),x_2=f(2),\ldots )$ and $\{\,f(1),f(2),\ldots\,\}$ are different objects: the first is an element of the Cartesian product and the second is a set of function values, so they cannot be equal.

However, with $f(i)=x_i \in X_i$, any permutation won't change the situation significantly, i.e. it doesn't matter if we consider an ordered tuple or a set of value. Again we have the problem with uncountable index sets. In case of $I=\mathbb{N}$ we have a nice ordering and can talk about tuples. But what to do if $I=\mathbb{R}\,?$ To match this general situation, direct products are defined without the necessity of an ordering, namely as functions $f$ instead of tuples. The distinction between e.g. $X_1\times X_2$ and $X_2 \times X_1$ isn't important, as long as we do not mix them: $\{\,f \, : \,\{\,1,2\,\} \longrightarrow X_1 \cup X_2\,|\,f(1)\in X_1 \wedge f(2)\in X_2\,\}$ avoids this confusion. Sure, in case $I=\mathbb{N}$ we normally list $(x_1,x_2,\ldots)$ since it is convenient and there is no need to artificially apply a permutation of the indices. But you are right, the formal definition doesn't consider any ordering.

For short: If $I$ has a natural ordering, we normally use it in our notation, if it has not, then we can't.

Another example is how we right functions in general. E.g. we are used to write $f(n)=\frac{1}{n}$ as $(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots)$ but at the same time would never try this for $f(x)=\frac{1}{x}\; , \;x\in \mathbb{R}_{>0}\,.$

 

[Periwinkle]

I was studying Group Theory on my own from a mathematics journal and got confused at some point where it defines Cartesian products, from binary one, say ($A × B$), to n-tuples one, say ($A_1 × A_2 × ... × A_n$). What confuses me when I tried to read it is that the definition made for infinite Cartesian product as shown below:
View attachment 233130
If I am correct that on the left-hand side $\prod_{i\in{I}}X_i$ by definition of Cartesian product corresponds to $X_1 × X_2 × ...$, which can be represented as in $(x_1,x_2,...)$. However, when I try to read and interpret the right-hand side, I fail to create those tuples $(x_1,x_2,...)$, but think of a set of $\{f_i\}$, which seemed to me absurd that how this set can turn into tuples.

I would like to know how to interpret all.

Someone rightly asks: Why is the union set operation in the above definition? When someone looks at A.A. Fraenkel, Y. Bar-Hillel, A. Levy, Foundations of Set Theory, on page 40, in the first footnote, the authors do not refer to the Descartes product in this general case, they use the name of the outer product. This points to the essence of the question.

Namely, for $n = 2$, the Descartes product can be easily defined by ordered pairs. $n = 3, n = 4$, and so on in any case. Also in axiomatic set theory. However, in the case of an infinite number of elements, sorting is much more doubtful. How can we sort the real numbers in order? The inventors of the logical and consistent system of set theory have encountered difficulties here, and the method described above was used to determine the outer product.

The above selection functions $f$ create selection sets. These selection functions correspond to the ordered pairs or ordered triples of the Cartesian product. The existence of the above outer product set itself is ensured by the subset axiom, indeed, this existing set may be the empty set. However, neither this subset axiom nor any other means can prove that the outer product set is not an empty set, even if none of the above $X_i$ set is empty. Considering that the ordered pair, the ordered triple, cannot be used - because we have developed another method for replacing them - the outer product does not have a proper and usual meaning, so this statement may not be so surprising. The Axiom of Choice enters here.