### 1 Answer

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- 8 months ago
First step is to find S6.

We do that by computing S1 through S5.

We have S1 = S0. So S1 = {0}

Then S2 = S1 U S2 = {0} U {0} = {0}

Knowing S1 and S2 it is analogous that any Sn = {0}

Therefore S6 = {0}

Now what is a power set?

The power set is the set of all subsets of S, including the empty set and S itself.

Let's take an example.

For S = {1, 2}

The power set of S is P(S) = { {}, {1}, {2}, {1, 2} }

So, in layman's terms, it's all the different combinations you can create with the elements of S.

Knowing this, the power set of S6 is:

P(S6) = { {}, {0} }

So, depending if you want to count the set of outer brackets, the number of opening braces is either 2 or 3

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