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I just found out that you can have $2^{2^n-1}$ different subsets, made from $n$ sets, using operations $\cup , \cap , \setminus$.

That is because when $n=2$, for example, you have 3 disjoint subsets: $A\setminus B, B\setminus A,A\cap B$ and then you can union those things together in $2^3-1$ different ways (plus the empty set) to get $8$ different unequal sets.

My question is, how to prove that the number of those disjoint subsets is $2^n-1$.

Pictorial definition of what disjoint subsets are:

enter image description here

In the picture above I have enumerated disjoint subsets of $2$ and $3$ sets. As you can see their number is $2^n-1$.

Pavel
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2 Answers2

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First of all, your question is not really well-defined, or put differently, the claim is incorrect in general. Consider for example $A = B = \emptyset$, in which there is only one set which you can get, namely $\emptyset$.

There are at least two ways in which we can address this problem. The first is to assume that the starting sets are "generic", that is, they don't enjoy any "special" properties. This is a vague description, but it can be formalized by saying that in the "Venn diagram" of the sets (which I describe below), all regions are non-empty.

The second way is to do everything formally. We think of $A$ or $A \setminus B$ not as a set but as a formal expression defined in an appropriate language. We then make two expressions equal if they are equal whenever you substitute actual sets. For example, $A = A \cap A$ and $A \setminus (A \setminus B) = A \cap B$. You can also define equality syntactically if you prefer.

Another problem with your claim is that it's not clear what you mean by "the sets have $2^n-1$ non-overlapping subsets". For example, $A \cap B$ and $A \setminus B$ are non-overlapping, and this is a maximal collection of non-overlapping subsets (in general). You mean something else, and again there are several possible interpretations. This time all interpretations I can think of only serve to complicate things and make them less intuitive, so I'll just explain what is meant by the claim, and you think of an appropriate interpretation.

The Venn diagram of the sets $A_1,\ldots,A_n$ consists of the $2^n$ regions which you can get by intersecting either $A_i$ or $\overline{A_i}$ for each $i$. That is, for each $i = 1,\ldots,n$ we choose $A_i$ or $\overline{A_i}$, and then compute the intersection of these $n$ sets. You can web-search "Venn diagram" to see pictorial representations.

You can prove by induction (exercise) that any set that you can obtain from (possibly multiple copies of) $A_1,\ldots,A_n$ using $\setminus$, $\cup$ and $\cap$ is a union of regions of the Venn diagram; indeed, a union which doesn't include the region $\overline{A_1} \cap \cdots \cap \overline{A_n}$. Moreover, you can easily get any of these $2^n-1$ regions individually (exercise), and these are your "$2^n-1$ non-overlapping subsets". In total, you can get any union of these $2^n-1$ regions (exercise), and so you can "reach" $2^{2^n-1}$ different expressions.

One possible formulation of your two claims are as follows. First, the maximal number of disjoint sets that can be formed from $A_1,\ldots,A_n$ is $2^n-1$ (under the caveats described in the second paragraph); this can be proved using the exercises mentioned above. Second, the number of different sets that can be formed (again, under the same caveats) is $2^{2^n-1}$, which again follows from these exercises.

Yuval Filmus
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  • Suppose every 2 sets among n sets have at least 1 element in common and 1 element not in common, is what I meant. I'm reading a book and it also doesn't define anything properly and I have to keep guessing what the authors mean.. But I'm amazed by how some people still understand what they mean. – Pavel Apr 15 '16 at 20:58
  • That's not enough. You have to assume that all regions (other than the outside region) in the Venn diagram are non-empty. – Yuval Filmus Apr 15 '16 at 21:03
  • I don't exactly understand what you mean by that, what do you mean by "regions"? – Pavel Apr 15 '16 at 21:06
  • You mean that all non-overlapping subsets have 1 element in common with the new set in my proof? – Pavel Apr 15 '16 at 21:07
  • The "regions" are the intersections of $A_i/\overline{A_i}$. You can think of them as the actual regions in the Venn diagram. There are $2^n$ of them, but the outside one $\overline{A_1} \cap \cdots \cap \overline{A_n}$ doesn't come into play here. – Yuval Filmus Apr 15 '16 at 21:08
  • I have no idea what you mean by "non-overlapping subsets". What you need is, for example when $n=3$, that all of $A_1 \cap A_2 \cap A_3$, $\overline{A_1} \cap A_2 \cap A_3$, $A_1 \cap \overline{A_2} \cap A_3$, $A_1 \cap A_2 \cap \overline{A_3}$, $\overline{A_1} \cap \overline{A_2} \cap A_3$, $\overline{A_1} \cap A_2 \cap \overline{A_3}$, $A_1 \cap \overline{A_2} \cap \overline{A_3}$ be non-empty. – Yuval Filmus Apr 15 '16 at 21:10
  • I will add to the question what I mean by non-overlapping subsets. – Pavel Apr 15 '16 at 21:11
  • Just added, is that clear? How Do I properly define that? – Pavel Apr 15 '16 at 21:16
  • The usual technical term is "disjoint sets". But your main claim, that there are $2^n-1$ disjoint sets that can be formed from $n$ sets using $\cap,\cup,\setminus$, still doesn't make much sense. You can also form less than $2^n-1$ disjoint sets if you wish, for example the set $A_1$ is "disjoint", the sets $A_1 \cap A_2 = A_1 \setminus A_2$ are disjoint, and so on. If you are just counting regions in the Venn diagram, then you're right that there are $2^n-1$ regions which are not the outside region. – Yuval Filmus Apr 15 '16 at 21:17
  • Yea, that's what I'm counting. Sorry, I will change the term to "disjoint subsets". I know that you can form less than that, so it's called "maximum number of disjoint subsets", I guess. – Pavel Apr 15 '16 at 21:19
  • In that case, you can prove your theorem from my claim that any expression that you get is a union of regions of the Venn diagram (exercise). – Yuval Filmus Apr 15 '16 at 21:22
  • You mean the theorem about number $2^n-1$ or the number $2^{2^n-1}$? The former I'm afraid I won't be able to write properly. The latter is easy because the number of binary numbers formed from $n$ bits is $2^n$. But I understand it intuitively. – Pavel Apr 15 '16 at 21:29
  • I mean the former, now that you have given it an explicit meaning. – Yuval Filmus Apr 15 '16 at 21:30
  • Is my proof bad then? To me it's obvious that if every region is supposed to be non-empty, then I will have to intersect the new set with $2^k-1$ of previous subsets and 1 part will have to be non-empty, too. I really don't get how I can do it more rigorously. But thanks a lot for your pointers, I need to learn to define things more formally. – Pavel Apr 15 '16 at 21:42
  • Your proof is ok at the intuitive level. – Yuval Filmus Apr 15 '16 at 21:43
  • Can you write it formally so I will accept the answer? This isn't homework or anything, I'm just reading a book on introductory set theory by Vereschagin (it's in Russian). So far I believe I can only prove things intuitively.. – Pavel Apr 15 '16 at 21:46
  • The main issue is that you haven't formulated your question rigorously. I tried to explain why in my answer. If you want anything more, I suggest studying for a math degree in a good university. – Yuval Filmus Apr 15 '16 at 21:49
  • Is it possible to reach a good level on your own? I'm an undergraduate now, finishing my last year of bachelor's degree.. – Pavel Apr 15 '16 at 22:03
  • In principle, but feedback from more experienced people can really help. – Yuval Filmus Apr 15 '16 at 22:32
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Proof by induction.

Suppose for $n=2$ it's true. (Obvious from Venn diagram with 2 sets).

Let's prove for $n=k+1$:

Suppose we have k sets, then we have $2^k-1$ disjoint subsets. Now we add another set. Then that set is supposed to be in intersection with $2^k-1$ of previous subsets and have a part that doesn't intersect with anything, hence we get $2*(2^k-1)+1=2^{k+1}-1$ subsets.

enter image description here

From picture above you can see how C is added to A and B.

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