In short: I look for a definition of a sum of any number of natural numbers in the terms of pure set theory. Until now, neither have I found such a definition in books, nor invented it by myself.

In details:

Let there be $n$ piles of apples on a table (${n}\in\mathbb{N}_{>0}$). Let $x_i$ be the number of apples in each pile (${x_i}\in\mathbb{N}_{>0}$, ${i}=1…n$). How to define the conception of “total number of apples on the table” through ${x_i}$, without using the operation of arithmetic addition?

All sources known to me reduce this conception to the arithmetic addition one way or another. But it seems not quite correct to me: addition doesn’t reflect the main point of the conception, but it only is one of the possible operations for calculating this “total number”. Besides that, the entity of “total number of apples on the table” exists regardless of the fact whether we perform any operations to calculate it.

Furthermore, addition is defined for two or more addends, while “total number of apples on the table” exists and is computable even if $n=1$.

I am interested in a definition in terms of pure set theory. Individual natural numbers (for example, $n$ and each of ${x_i}$) can be defined, e.g. as finite ordinals. I look for a definition of “total number” also in the context of set theory (e.g. as a result of unions, intersections and other set operations).

Is this possible?

  • 2
    $\begingroup$ How do you even distinguish between "sum" and "addition"? By definition addition means the task of finding a sum (or in other words, "addition" is the relation between the summands and the sum), and a sum means what you get out of an addition. $\endgroup$ – hmakholm left over Monica Dec 23 '16 at 13:50
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    $\begingroup$ See the post : how is addition defined for the set-theoretic definition of sum. $\endgroup$ – Mauro ALLEGRANZA Dec 23 '16 at 13:51
  • $\begingroup$ @Henning, this is why I say “total number of apples on the table”, not “sum” in the body of my question. In the title, I used the word “sum” only for the sake of brevity. Of course, sum is the result of addition, but “total number” exists even if we do not perform any additions. $\endgroup$ – Hydrochoerus Hydrochaeris Dec 23 '16 at 14:05
  • $\begingroup$ How do you define the number of apples? what is a number? if a number is an ordinal (i.e., a transitive and $\in$-well-ordered set), and if your piles $I$ are well-ordered too, then the total number could be the ordinal $\alpha=\{<i,x_i>:i\in I,\ x_i\in n_i\}$ where, for $i\in I$, $n_i$ is the number of apples in pile $i$, and the well-ordering of $\alpha$ is: $<i,x><<j,y>\iff i<j\lor(i=j\land x<y)$. $\endgroup$ – gniourf_gniourf Dec 23 '16 at 14:05
  • $\begingroup$ Thank you @Mauro, but in that post I see definitions of addition operation, whereas I look for a definition of “total number” without performing the addition. $\endgroup$ – Hydrochoerus Hydrochaeris Dec 23 '16 at 14:09

If you have an indexed family of cardinalities $(\kappa_i)_{i\in I}$, then you can define the sum of the $\kappa_i$s to mean any cardinality $\lambda$ where

  • There is a family of sets $(A_i)_{i\in I}$ such that ...
  • For each $i$ it holds that $|A_i|=\kappa_i$, and
  • The $A_i$s are pairwise disjoint, and
  • $\lambda = \left| \bigcup_{i\in I} A_i\right|$.

I will leave it to you to prove that

  1. Every family $(\kappa_i)_{i\in I}$ has at least one sum (easy).

  2. Every family $(\kappa_i)_{i\in I}$ has as most one sum (fairly easy if you assume the axiom of choice; but not provable in ZF. It appears to be unknown whether it implies the axiom of choice).

  3. The sum of a finite family of finite numbers is finite (possibly hard, depending on how you define "finite", and how purist you are about not giving binary addition any special treatment).

  • $\begingroup$ Thank you, @Henning. But if the sets are natural numbers, I don’t understand how to make them “pairwise disjoint”. For example, if we are to define the “total number” of 2, 3, 3 and 5—then your λ is a cardinality of a ∪ of what? $\endgroup$ – Hydrochoerus Hydrochaeris Dec 23 '16 at 14:39
  • $\begingroup$ @Displayname: You could for example choose $$\{11,12\}\cup \{21,22,23\}\cup\{31,32,33\}\cup\{41,42,43,44,45\}$$ There are many possibilities, and step 2 of the part I'm leaving to you is to show that it doesn't matter which of them you use. $\endgroup$ – hmakholm left over Monica Dec 23 '16 at 14:42
  • $\begingroup$ In your example, you implicitly establish an order between sets (as it were “marking” one of them with index 1, another with 2 etc.). I understand the idea of replacing each set with another set of the same cardinality so that all sets become disjoint. But I don’t understand how to construct this “replacing” without ordering the sets, implicitly or explicitly. $\endgroup$ – Hydrochoerus Hydrochaeris Dec 23 '16 at 15:14
  • $\begingroup$ @Displayname: I need to write down my example in some particular order due to the left-to-right nature of writing -- but in the actual definition in the answer you're perfectly free to use an index set $I$ that does not come with a preferred ordering. $\endgroup$ – hmakholm left over Monica Dec 23 '16 at 15:16

We use the following definition (see this wikipedia link),

Definition: A set $S$ is said to be finite if it can be given a total ordering which is well-ordered both forwards and backwards. That is, every non-empty subset of S has both a least and a greatest element in the subset.

We define the operator $\Gamma$ to map any $(A,\le)$ well-ordered set as follows:

$ \Gamma(A,\le)=\begin{cases} \bigr(A \setminus \{\text{min}(A)\},\,\rho_{\le} \setminus \{ \text{min}(A)\}\times A \bigr) &\text{where } A \ne \emptyset \\ (\emptyset,\emptyset)&\text{otherwise} \end{cases} $

If presented with a finite family of finite sets $(A_i)_{i\in I}$ that are pairwise disjoint, then put a total ordering $\rho_{\le}$ on the union

$\quad A = \bigcup_{i\in I} A_i$

so that $A$ is well-ordered both forwards and backwards.

Define by recursion a function operating on the finite ordinals $\omega$ via

$\quad g(\emptyset) = (A, \le)$

$\quad g(\alpha \cup \{\alpha\}) = \Gamma\bigr(g(\alpha)\bigr)$

There exists an ordinal $\kappa$ such that $g(\kappa) = (\emptyset,\emptyset)$. Let

$\quad \mathcal E = \{ \kappa \in \omega \mid g(\kappa) = (\emptyset,\emptyset)\}$

If $\mathcal E = \omega$ then the sum of the $A_i$ is equal to $\emptyset$ (i.e. $0$).

Otherwise the sum is the ordinal preceding $\text{min}(\mathcal E)$.

The OP has to confirm that the sum so specified does not depend on how the set $A$ is well-ordered.


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