Define an operator $\mathop{\vec{\bigcup}}$ as follows:

Definition. Whenever $A$ is an $I$-indexed family of sets, where $I$ is a totally-ordered set, we have $$\mathop{\vec{\bigcup}}_{i \in I} A_i = \bigcup_{j<i} A_j$$

For example,

$$\mathop{\vec{\bigcup}}_{i \in \mathbb{N}} \{i\} = \{0,\ldots,i-1\}.$$

Convention. For the purposes of this question, lets refer to $\mathop{\vec{\bigcup}}$ as the serialization of the operator $\bigcup$.

We can serialize other operators, too, of course, using the exact same definition as above. For example:

$$\mathop{\vec{\sum}}_{i \in \mathbb{N}} i = \frac{1}{2} i(i-1)$$

And of course, this features in the definition of an infinite sum:

$$\sum_{i \in \mathbb{N}} a_i = \lim_{i \rightarrow \infty} \mathop{\vec{\sum_{i \in \mathbb{N}}}}a_i$$

In this context, the entity $\mathop{\vec{\sum_{i \in \mathbb{N}}}}$ could be referred to as the 'partial sum operator.'

Something quite similar seems to happen in calculus; when we write something like $$\int_5 2x dx = x^2-5^2$$ we're using the ordered structure of the real line, together with our ability to integrate over subsets of the real line, together with a distinguished basepoint, namely $5$, to "serialize" the Riemann integral (or Lebesgue integral, for that matter) with respect to $5$. FTC then tells us that, if certain hypothesize are met, the serialization of the Riemann integral is an inverse to differentiation.

Indeed, backing up a bit, lets define that:

Definition. If $A$ is an $\mathbb{N}$-indexed family of sets, then

$$\mathop{\mathrm{disj}}_{i \in \mathbb{N}}A_i = A_i \setminus \mathop{\vec{\bigcup}}_{i \in I} A_i.$$

This codifies the "disjointification trick" from probability theory and measure theory. Given the connection between differentiation and integration, I guess it makes sense to think of $\mathop{\mathrm{disj}}$ as playing the role of differentiation in the world of boolean algebra. We have a kind of fundamental theorem of calculus, namely:

$$ \mathop{\vec{\bigcup}}_{i \in \mathbb{N}} \mathop{\mathrm{disj}}_{i \in \mathbb{N}} A_i = A_i,$$

and it seems to be the case that $\mathop{\mathrm{disj}}_{i \in \mathbb{N}} A_i$ produces the "smallest" (in the sense of $\subseteq$) sequence making the above formula true; I guess they're order-theoretic adjoints or something.


Q0. Is there accepted terminology for what I'm calling "serialization"?

Q1. Is there any existent theory surrounding this concept?

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    $\begingroup$ It seems to me that the notation could be improved, since the subscript $i$ gives $i$ the appearance of a bound variable, with the $A_i$ appearing under its scope, as in $\vec\bigcup_{i\in I}A_i$, but that is not what you mean, since you indicate that the output is something that depends on $i$. One can easily manufacture ambiguous expressions with this kind of notation. It seems to me that what $\vec\bigcup$ should really output is a function from $I$ to those instances. $\endgroup$ – Joel David Hamkins Jul 2 '17 at 14:47
  • $\begingroup$ @JoelDavidHamkins, I think what you're saying is equivalent to the position that $\frac{\partial}{\partial x} E$ could be improved by changing it to $(D(\lambda x.E))(x')$ or some such, which I don't really agree with. But I'm happy to hear you out. Can you give more details about these ambiguous expressions? What's a simple example of such a thing? $\endgroup$ – goblin GONE Jul 2 '17 at 14:50
  • $\begingroup$ Suppose I want to evaluate the $i=2$ instance of your expression $\vec\bigcup_{i\in I}A_i$. Can I write $\vec\bigcup_{i\in I}A_2$? I don't think so. Suppose that I re-index by one; what does $\vec\bigvee_{i\in \mathbb{Z}}A_{i+1}$ mean, as opposed to $\vec\bigvee_{i\in I}A_i$? $\endgroup$ – Joel David Hamkins Jul 2 '17 at 15:01
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    $\begingroup$ And in some cases where you actually do care about the order, the notion is ill-behaved. The limit superior of a sequence is the limit (usual order) of the "anti-serialization" (serialization under the dual order) of the sup-operator. Without seeing some interesting argument applying to all serializations, I don't see why one should expect a general theory. $\endgroup$ – Michael Greinecker Jul 2 '17 at 15:18
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    $\begingroup$ You should really change the notation here, is the operator aplied to the whole family of sets? or is it just applied to the sets whose index is lower than some fixed $i$ ? It makes it unnecessarily more difficult to understand. $\endgroup$ – Max Jul 2 '17 at 15:26

Your concept is similar to (but not exactly the same as) the concept of diagonal union, defined for ordinals $\kappa$ and sets $A_\alpha\subset\kappa$ for $\alpha<\kappa$ as follows:

$$\mathop{\bigtriangledown}_{\alpha<\kappa}A_\alpha\quad=\quad\{\beta<\kappa\mid \beta\in\bigcup_{\alpha<\beta}A_\alpha\}.$$

The dual notion of diagonal intersection is defined by

$$\mathop{\bigtriangleup}_{\alpha<\kappa}A_\alpha\quad=\quad\{\beta<\kappa\mid \beta\in\bigcap_{\alpha<\beta}A_\alpha\}.$$

These operations are used all over set theory, principally because the club filter on an uncountable regular cardinal is closed under diagonal intersection. A normal filter, a basic concept, is one that is closed under diagonal intersection. There is a huge theory developing and surrounding this, which begins in any good set theory text.

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  • $\begingroup$ It seems that the idea of serialization works generally best for well-ordered index sets. Transfinite recursion is basically constructing a function that has the structure of a serialization. $\endgroup$ – Michael Greinecker Jul 2 '17 at 15:28

I think more care is needed. You could consider serialization as a mapping from $A*I$ to $A*I$, where I am using this notation to denote a kind of power of A, but not a Cartesian product, as order is important. An example of how powerful and confusing it can be is to take A and I both to be a modification of the rational numbers. Do you want the result to look like an infinite dimensional vector space over the rational field, or do you want the result to look like the real numbers as constructed by a Dedekind completion of the rational numbers?

Gerhard "Or Just Operate Infinitely Often?" Paseman, 2017.07.02.

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