Let's analyze each part of the question step-by-step, focusing on calculating the number of elements in the power sets of the given sets.

a) $\mathcal{P}(\{a, b, \{a, b\}\})$

The power set $\mathcal{P}(S)$ of a set $S$ is the set of all subsets of $S$. The set in question here is:

$\{a, b, \{a, b\}\}$

This set has three elements: $a$, $b$, and the set $\{a, b\}$.

The power set of a set with $n$ elements has $2^n$ elements. Since this set has 3 elements, the power set will have:

$2^3 = 8 \text{ elements.}$

b) $\mathcal{P}(\{\emptyset, a, \{a\}, \{\{a\}\}\})$

The set in question is:

$\{\emptyset, a, \{a\}, \{\{a\}\}\}$

This set has four distinct elements: $\emptyset$ (the empty set), $a$, the set $\{a\}$, and the set $\{\{a\}\}$.

The power set of a set with $n$ elements has $2^n$ elements. Since this set has 4 elements, the power set will have:

$2^4 = 16 \text{ elements.}$

c) $\mathcal{P}(\mathcal{P}(\emptyset))$

First, let's compute $\mathcal{P}(\emptyset)$. The empty set $\emptyset$ has exactly one subset, itself. So:

$\mathcal{P}(\emptyset) = \{\emptyset\}$

This set has one element, the empty set. Now, we compute the power set of $\{\emptyset\}$, which contains 1 element. The power set of a set with 1 element contains:

$2^1 = 2 \text{ elements.}$

Thus, $\mathcal{P}(\mathcal{P}(\emptyset))$ has 2 elements.

Final Answer:

• a) 8 elements
• b) 16 elements
• c) 2 elements

Would you like further details on any of the steps? Let me know!

Here are 5 related questions you might explore:

1. What is the power set of $\mathcal{P}(\{a, b\})$?
2. How many elements are in the power set of a set with 5 elements?
3. Can a power set of a set ever have fewer elements than the original set?
4. How does the power set operation behave when applied repeatedly (e.g., $\mathcal{P}(\mathcal{P}(\mathcal{P}(S)))$)?
5. What are the elements of $\mathcal{P}(\mathcal{P}(\{a\}))$?

Tip: The number of elements in a power set grows exponentially with the number of elements in the original set!