Counting subsets.

Question

I know we can use induction to prove statements about counting various types of subsets of a set and that we have to apply the same idea to count the number of pairs of disjoint subsets of a set.

I am given with the following definitions:

• $(S_{n})$: Let $n∈N$. We define the set $S_{n} = \{0, 1, . . . , n−1\}$ (with $S_{0}$ obviously $= ∅$). Note that $|S_{n}| = n$.
• $(DP_{n})$: Let $n∈N$. We define the set of disjoint subset pairs of $S_{n}$ as: $$ DP_{n}=\{(A, B) \ | \ A, B ⊆ S_{n} \ \text{and} \ A ∩ B = ∅\} $$

From this I've inferred for $n = 0$, $S_{n} = ∅$, and $DP_{n} = \{(∅, ∅)\}$. For $n = 3$, $S_{n}= \{0, 1, 2\}$, some elements of $DP_{3}$ are $(\{1\}, \{0, 2\})$ and $(\{0, 1, 2\}, ∅)$ and that order matters in tuples so the pair $(\{0, 2\}, \{1\})$ in $DP_{3}$, and is considered different from $(\{1\}, \{0, 2\})$. Please correct me if any of these assumptions is wrong.

(a) What are $DP_{1}$ and $DP_{2}$ explicitly in set notation?

(b) Find a formula for the size of $DP_{n}$ (in terms of $n$), and then prove that your formula is correct for all values of $n$ ($\forall n \in \mathbb{N}$).

Answer 1

This is what I did for (b):

Let $n \in \mathbb{N}$. Then for the base case, $n=0$: \begin{align*} L.H.S.=|DP_0|=1\\ R.H.S.=3^0=1 \end{align*} So $P(0)$ holds.

Assuming that $P(k)$ holds for $n=k$, we have $|DP_k|=3^k$ And we need to prove $|DP_{k+1}|=3^{k+1}$ for $n=k+1$.

Now, \begin{align*} |DP_{k+1}| &=|DP_k| \cup |\{ \text{the pairs of elements in the set $S_k$} \}|\\ &=3^k +2.3^k && \text{(The $2$ is to account for the pairs occurring}\\ &= 3^k(1+2) && \text{in the subsequent set being added since order}\\ &= 3^k.3 && \text{does matter in tuples.)}\\ &= 3^{k+1} \end{align*} which is what we wanted to show.

$\therefore P(k+1)$ holds.

Answer 2

From this I've inferred for $n = 0$, $S_{n} = ∅$, and $DP_{n} = \{(∅, > ∅)\}$. For $n = 3$, $S_{n}= \{0, 1, 2\}$, some elements of $DP_{3}$ are $(\{1\}, \{0, 2\})$ and $(\{0, 1, 2\}, ∅)$ and that order matters in tuples so the pair $(\{0, 2\}, \{1\})$ in $DP_{3}$, and is considered different from $(\{1\}, \{0, 2\})$.

Yes, this is correct.

(a) Start from $S_0=\{\}$, $DP_0 = \{(\emptyset, \emptyset)\}$

Add in the pairs due to new the element $0$ in $S_1=\{0\}$: $$ DP_1 = DP_0 \cup \{(\emptyset,0), (0,\emptyset) \} $$

Similarly, the pairs due to the new element $1$ in $S_2=\{0,1\}$: $$ DP_2 = DP_1\cup \{(0,1),(1,0),(1,\emptyset),...,(\{0,1 \},\emptyset) \} $$

Note that the recursive structure of $DP_n$. This is key for an inductive proof; when you attempt it, you will see that: $|DP_n|=|DP_{n-1}|+2\cdot 3^{n-1}$

(b) This is the same formula as provided in the question linked by Ross Millikan; although I will explain a little more here.

You could pick any $m$ elements out of $n$ to form a subset $A$ (the number of ways to do this is given by the binomial coffifient); then to form the other disjoint subset $B$, you can pick any combination of the remaining $n-m$ elements. There are $2^{n-m}$ possibilities.(Why?)

To conclude, for each subset $A$ of size $m$(of which there are $\binom{n}{m}$), there are $2^{n-m}$ disjoint subsets ($B$); and $m$ can vary from zero(when $A$ is the empty set) all the way up to $n$(when $B$ is the empty set). Thus giving us:

$$ \sum_{m=0}^{n}\binom{n}{m}2^{n-m} = \sum_{m=0}^{n}\binom{n}{m}2^{n-m}\cdot 1^m = 3^n $$ The final equality above follows from the binomial theorem: $$ (a+b)^n =\sum_{m=0}^n a^{n-m}\cdot b^m $$

The above is a proof enough, but the way the question is phrased it looks like it's asking you to rigorously prove that that $DP_n$ has $3^n$ elements by using induction on $n$.

Hint: Assume $DP_n$ has $3^n$ elements and count the number of subset pairs in $DP_{n+1}$that contain the new element $n$ from $S_{n+1}=\{0,1,2...n\}$. (Like in (a), above.)

Read the combinatorial proof of Pascal's identity for inspiration if required.

Hint #2: The number of subsets $A$ of size $k$ containing the new element $n$ from $S_{n+1}=\{0,1,2...n\}$ is $\binom{n-1}{k-1}$(since $n$ is already in $A$, only $k-1$ more can be picked); and $k$ can vary from $1$ to $n$. Also, for each such $A$, you can pick any combination of elements from the remaining $n-k$ elements to get $B$. In addition to that, for each $(A,B)$ you also have $(B,A)$; remember to account for that.