`n{P(P(P(varphi)))}=`

Options

A. 6
B. 8
C. `2^(3-1)`
D. `2^(5-1)`

Answer

To solve the problem \( n{P(P(P(\varphi)))} \), we will break it down step by step.

### Step 1: Understanding the Problem
We need to find the cardinality of the power set of the power set of the power set of the empty set \( \varphi \). The notation \( P(X) \) denotes the power set of set \( X \), which is the set of all subsets of \( X \).

### Step 2: Finding \( P(\varphi) \)
The empty set \( \varphi \) has no elements. The power set of the empty set is:
\[ P(\varphi) = \{ \varphi \} \]
This means the power set of the empty set contains one element, which is the empty set itself. Therefore, the cardinality of \( P(\varphi) \) is:
\[ |P(\varphi)| = 1 \]

### Step 3: Finding \( P(P(\varphi)) \)
Now, we need to find the power set of \( P(\varphi) \):
\[ P(P(\varphi)) = P(\{ \varphi \}) \]
The set \( \{ \varphi \} \) has one element. The power set of a set with one element contains two subsets: the empty set and the set itself. Thus:
\[ P(P(\varphi)) = \{ \varphi, \{ \varphi \} \} \]
The cardinality of \( P(P(\varphi)) \) is:
\[ |P(P(\varphi))| = 2 \]

### Step 4: Finding \( P(P(P(\varphi))) \)
Next, we find the power set of \( P(P(\varphi)) \):
\[ P(P(P(\varphi))) = P(\{ \varphi, \{ \varphi \} \}) \]
The set \( \{ \varphi, \{ \varphi \} \} \) has two elements. The power set of a set with two elements contains four subsets:
1. The empty set
2. The set containing only \( \varphi \)
3. The set containing only \( \{ \varphi \} \)
4. The set containing both \( \varphi \) and \( \{ \varphi \} \)

Thus:
\[ P(P(P(\varphi))) = \{ \varphi, \{ \varphi \}, \{ \{ \varphi \} \}, \{ \varphi, \{ \varphi \} \} \} \]
The cardinality of \( P(P(P(\varphi))) \) is:
\[ |P(P(P(\varphi)))| = 4 \]

### Step 5: Final Calculation
Now, we need to find the cardinality of the cardinality of \( P(P(P(\varphi))) \):
\[ n{P(P(P(\varphi)))} = |P(P(P(\varphi)))| = 4 \]

### Conclusion
The final answer is:
\[ n{P(P(P(\varphi)))} = 4 \]