Given a set $A$

I would like to know if already exist a math definition for a Set $L$ being:

$L$="the smaller cardinal set from a choice between a set and its complement"

i.e a set with this properties

$L=A$ , if $|A|$ $\lt$ $|\overline A|$

$L=\overline A$ , if $|A|$ $\gt$$|\overline A|$

and $L=A$ or $L=\overline A$ (indistinct when tie) , if $|A|$ = $|\overline A|$

I just wonder if it has any known use or name, perhaps is too trivial to have a name, but that's the question. thanks


$A$ is a subset of a countable set $X$.

UPDATE 2: a comment about this function

As I told before the operation is very simple to have a name, but just for reference, or fun, I would like to share with you a comment about the source, I call it the Law operation to the set A, L(A) (it could be called the legislation function or rule operation, etc..). Why? because in every legislation process there are admissible and inadmissible things, legal and illegal, permitted and prohibited, possible and impossible and so on.. for example in physics there are laws of physics, that include and exclude certain possibilities, the fact is that we never call it laws when the cardinal of excluded things are bigger than the included, because if were so, we would just legislate the complement, i.e. fewer laws to legislate the same. L could be from Leibniz too, because it's his idea that "when a rule is very complex, what is conformable to it is seen as irregular", so I wanted to reduce that concept, to the minimum complexity taking into account only the number of facts that the law regulates.

  • 1
    $\begingroup$ The set $A$ should be a subset of some particular set, say $X$. Otherwise it is not so clear what $\overline A$ is. The cardinality of $\overline A$ is not well defined unless we know what $X$ is. I am not aware of a name for this function. $\endgroup$ – Jay Jan 19 '12 at 1:31
  • $\begingroup$ @Jay thanks for the comment, I've added it $\endgroup$ – Hernán Eche Jan 19 '12 at 2:28
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    $\begingroup$ I do not know any standard operator that would do this; you can simply define one, though if you really want it to be an operator, you have to pick whether you'll take $A$ or $\overline{A}$ when $A$ is both infinite and coinfinite. $\endgroup$ – Arturo Magidin Jan 19 '12 at 5:46
  • $\begingroup$ @ArturoMagidin that's a tip for freedom, thanks $\endgroup$ – Hernán Eche Jan 19 '12 at 11:36

Firstly, note that subsets of countable sets are either finite or countable namely the set of possible values is $\mathbb N\cup\{\aleph_0\}$ with the natural ordering of $\mathbb N$ and $\aleph_0$ being the maximum.

This order is linear, therefore such operation is well defined as every two cardinals are comparable.

Secondly, most set theory is in a general settings without taking such $X$. So if such operator was common it would have to include $X$ as the parameter. Since I am writing this, you may guess that the answer is negative and such operation is not common in mathematics. This means that you simply have to define it when you need to use it, this is a perfectly fine procedure to define ad-hoc operations when you need them (just remember to show that they are well defined. As I remarked above, in this case it is).

So you can simply write something like this:

Define $L(A)=\begin{cases} A & |A|\le|X\setminus A|\\ X\setminus A & |X\setminus A|<|A|\end{cases}$, since $X$ is countable $A$ is either finite or countable and therefore we can compare these cardinalities and find the minimal one.


The general case (i.e. when $X$ is any given set) requires extra assumptions, namely that every two cardinalities are comparable. If you are in such settings which do not allow such assumptions it may not be well defined when $X$ is uncountable.

One remark on the name, names of operations are often forgotten and we are only left with the notation. In this case, $L(A)$ is quite the busy notation in set theory, so if you plan on using it in the context of set theory itself I suggest a different notation. If, on the other hand, you plan on using it in different contexts I suppose it can be fine.

  • $\begingroup$ +1 Good for highlight the comparable cardinals restriction and the relative complement notation, there is still a detail to solve, in your L(A) bracket I think it would remains to specify that it could be $L(A) = A$ even when $| X\setminus A |$ = $| A |$, but that's not specified. Perhaps that "detail" make things being complex because there is not any criterion of choice, or perhaps I could simply add $L=A \vee L= X\setminus A$ when $| X\setminus A |$ = $| A |$, but I doubt that it keep being "well defined", then an arbitrary choice of L(A)=A in the definition could be enough $\endgroup$ – Hernán Eche Jan 19 '12 at 12:47
  • $\begingroup$ @Hernan: Note that I defined $L(A)=A$ when $|A|\le|X\setminus A|$, which includes the case of equal cardinalities. $\endgroup$ – Asaf Karagila Jan 19 '12 at 12:51

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