# How many curly brackets of the empty powerset?

The question is: How many curly brackets are there in the following, if $\varnothing$ counts as $\{ \}$ (=1 curly brackets)

$$\wp^5(\varnothing)$$

I calculated $p^2$, which was $4$ curly brackets, but at $p^3$: $$\wp^3(\varnothing)$$

I found out it has 8 curly brackets (still have to double check). But I'm really confused at how I can calculate this till the power of 5.

Thank you!

• Try to calculate $\mathcal{P}^4(\emptyset)$. It might be a bit tiring, but you should start to see a pattern emerge. – EuYu Jun 20 '14 at 23:33
• $\pe$ is the Weierstrauss elliptic function, $\mathcal{P}$ is the power set. – Rene Schipperus Jun 20 '14 at 23:33
• I wrote out $\mathcal{P}^{3}(\emptyset)$ and got $11$ pairs of curly brackets. You might want to double check that. – JimmyK4542 Jun 21 '14 at 0:21
• @Rene: I have seen from time to time that some people use $\wp$ for the power set. (The code is \wp by the way.) – Asaf Karagila Jun 21 '14 at 0:39
• Yeah I was trying to change it when the site went down. I think $\wp$ should be reserved for the elliptic function and a different symbol for power set should be used. – Rene Schipperus Jun 21 '14 at 0:56

Let $b_n$ be the number of elements in $\mathcal{P}^{n}(\emptyset)$. Let $a_n$ be the number of curly brackets in $\mathcal{P}^{n}(\emptyset)$.

Clearly, $b_n = 2^{b_{n-1}}$. Therefore, $b_0 = 0$, $b_1 = 1$, $b_2 = 2$, $b_3 = 4$, $b_4 = 16$, and $b_5 = 65536$.

Each element of $\mathcal{P}^{n-1}(\emptyset)$ will be in half of the subsets of $\mathcal{P}^{n-1}(\emptyset)$, i.e. half of the elements of $\mathcal{P}^{n}(\emptyset)$. Thus, each element of $\mathcal{P}^{n-1}(\emptyset)$ gets written down $\tfrac{b_n}{2}$ times in $\mathcal{P}^{n}(\emptyset)$.

This requires a total of $\tfrac{b_n}{2}(a_{n-1}-1)$ pairs of curly brackets.

Then, we need one pair of curly brackets for each element of $\mathcal{P}^{n}(\emptyset)$, (so $b_n$ of them), as well as one more pair for the outermost curly brackets in $\mathcal{P}^{n}(\emptyset)$.

Therefore, this is a total of $a_{n} = \tfrac{b_n}{2}(a_{n-1}-1) + b_n + 1 = \tfrac{b_n}{2}(a_{n-1}+1)+1$ curly brackets.

Now, crank out the recursion to get: $a_0 = 1$, $a_1 = 2$, $a_2 = 4$, $a_3 = 11$, $a_4 = 97$, $a_5 = 3211265$.

So, we need $3211265$ pairs of curly brackets to write out $\mathcal{P}^{5}(\emptyset)$.

• Thank you! I really spent 1 hour studying your answer and just got a eureka moment and understand it myself. Thank you! – user3125591 Jun 21 '14 at 12:26