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The usual definition of an arbitrary Cartesian product is: if $\{X_i\}_{i \in I}$ is an indexed family of sets, $\prod_{i \in I}X_i$ is defined to be the set of all functions $x:I \rightarrow \bigcup_{i \in I} X_i$ such that $x(i) \in X_i$ for all $i\in I$. But I believe that this definition cause some intuitive properties of the product to be false.

The main problem is that, every time $\{ X_i\}_{i\in I}$ and $\{ Y_i\}_{i\in I}$ are distinct families such that $\bigcup_{i \in I} X_i \neq \bigcup_{i \in I} Y_i$, then, rigorously, no function of the form $x:I \rightarrow \bigcup_{i \in I} X_i$ can be of the form $x:I \rightarrow \bigcup_{i \in I} Y_i$, because for two functions to be the same, they must have the same Codomain. This observation shows that, in this case, $(\prod_{i\in I} X_i) \bigcap (\prod_{i\in I} Y_i)= \emptyset$, and a consequence of this is that:

$$X_i \subseteq Y_i \ \ \forall i \in I \not \Rightarrow \prod_{i \in I} X_i \subseteq \prod_{i \in I} Y_i$$

(you can see this if you let the $X_i$ be proper subsets of the $Y_i$, for example).

I acknowledge that this problem is usually ignored for practical purposes, but I would like to know if there is a way of fixing this by changing something in the definition of the product (or if I am making any mistakes in my reasoning). Thanks in advance.

ConcreteSwanDive
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  • https://math.stackexchange.com/questions/1087412/what-is-a-function https://math.stackexchange.com/questions/1917581/how-do-i-define-exactly-what-a-function-is https://math.stackexchange.com/questions/871553/functions-with-different-codomain-the-same-according-to-my-book – Asaf Karagila Jul 31 '20 at 10:04

1 Answers1

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for two functions to be the same, they must have the same Codomain

This presupposes a particular notion of "function" which is not the one used in set theory. In set theory, a function is defined as a set of ordered pairs - with no explicit codomain. Phrased another way, there is no difference between a function $f$ and its graph $\{\langle a,b\rangle: f(a)=b\}$.

And it's this set-theoretic definition of "function" which is used when $\prod_{i\in I}X_i$ is defined as "the set of functions with domain $I$ and codomain $\bigcup_{i\in I}X_i$ sending each $i\in I$ to some element of $X_i$."

Of course, having an explicit codomain can be quite useful. So we can also talk about more complicated objects, pairs $\langle f, A\rangle$ where $f$ is a function in the above "weak" sense and $A$ is some set containing each right coordinate of an element of $f$ (a valid choice of codomain). In set theory there isn't a special term for these, since the "codomain-free" definition is generally more useful, but I've heard terms like "function-with-codomain" used here. But when we use the appropriate definition of function, the issue you point to does not arise.

Noah Schweber
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  • I wouldn't like being off topic, but in my algebra class we've actually given a different definition for a function(in general relation) and its graph. We defined a relation as a couple $R=(I \times J, G)$, where G is a subset of the power set of $I \times J$, and it's called graph. Is this a correct(alternative) approach? – Kandinskij Jul 31 '20 at 04:15
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    @Eureka Oh, well in that case it's not ideal to define $\Pi_{i\in I}X_i$ as "the set of functions with domain $I$ and codomain $\bigcup_{i\in I}X_i$ sending each $i\in I$ to some element of $X_i$" for exactly the reason you mention. One option is, as you suggest, sweeping the issue under the rug, but I really think this is a situation where the Cartesian product should be defined as a set of functions in the "codomain-free" sense above. – Noah Schweber Jul 31 '20 at 04:37
  • I'm not the one that asked the question – Kandinskij Jul 31 '20 at 04:47
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    @Eureka Whoops, my apologies. (My mathematical comment stands though.) – Noah Schweber Jul 31 '20 at 05:34
  • Thank you @NoahSchweber! I was not aware of this "codomain-free" definition. But, interestingly enough, I believe that even if you, in some sense, specify the codomain, you could still have this "liberty from codomain". If you define the relation $R$ as a subset of $I \times J$ and $J \subset K$, then $R$ is also a subset of $I \times K$ (which is true as we are dealing with the finite product here). It only remains to check that this subset is still a function. – ConcreteSwanDive Jul 31 '20 at 14:44