1. PPPP(empty set)

So I need to find $\mathcal{P} \mathcal{P} \mathcal{P} \mathcal{P} (\phi)$.

I start with $\phi$.

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

$\mathcal{P}\mathcal{P}(\phi)=\{ \phi, \{ \phi \} \}$

$\mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$

(After $\mathcal{P} (\phi)$, i'm just taking the union of the sets before it. Therefore (writing the elements as a list)):

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

$\phi$
$\{ \phi\}$
$\{ \phi, \{ \phi \} \}$
$\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$

$\}$

To me this makes sense since i'm constructing the natural numbers:

$\phi=0$
$\mathcal{P} (\phi)=\{ \phi \} =1$
$\mathcal{P} \mathcal {P} (\phi)=\{\phi, \{ \phi \} \}=2$ etc

However, my tutor wrote that I should have 16 elements in $\mathcal {P} \mathcal{P}\mathcal{P}\mathcal{P}(\phi)$. How is he getting all these extra elements?

2. Originally Posted by Showcase_22
So I need to find $\mathcal{P} \mathcal{P} \mathcal{P} \mathcal{P} (\phi)$.
I start with $\phi$.
$\mathcal{P}(\phi)=\{ \phi\}$
$\mathcal{P}\mathcal{P}(\phi)=\{ \phi, \{ \phi \} \}$
$\mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$
To me this makes sense since i'm constructing the natural numbers:
However, my tutor wrote that I should have 16 elements in
You missed a set at the third level.
$\mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{\{ \phi \}\},\{ \phi, \{ \phi \} \} \}$
The power set of a set of $n$ has $2^n$ members. That what your tutor means.

That said, that is not the way the natural numbers are defined.
This is
$0=\emptyset$
$1=\{0\}=\{\emptyset\}$
$2=\{0,1\}$
$3=\{0,1,2\}$
$.\quad \vdots$

3. Since I have already started writing...

If $|A|=n$, then $|\mathcal{P}(A)|=2^n$. Therefore, indeed, $|\mathcal{P}^4(\emptyset)|=16$.

Originally Posted by Showcase_22
$\mathcal{P}\mathcal{P}\mathcal{P}(\phi)=\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$
You need one more element: $\{\{\emptyset\}\}$.

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

$\phi$
$\{ \phi\}$
$\{ \phi, \{ \phi \} \}$
$\{ \phi , \{ \phi \}, \{ \phi, \{ \phi \} \} \}$

$\}$
If we denote the elements of $\mathcal{P}^3(\emptyset)$ as follows: $a_1=\emptyset$, $a_2=\{\emptyset\}$, $a_3=\{\{\emptyset\}\}$, $a_4=\{\emptyset,\{\emptyset\}\}$, then your list can be written as follows:

$\emptyset$, $\{a_1\}$, $\{a_1,a_2\}$, $\{a_1,a_2,a_4\}$.

It is missing 12 sets, including $\{a_2\}$, $\{a_4\}$, $\{a_1,a_4\}$, $\{a_2,a_4\}$, as well sets that contain $a_3$.

To me this makes sense since i'm constructing the natural numbers
Though you can encode natural numbers in this way, a standard construction (von Neumann ordinals) defines the successor of $x$ as $x\cup\{x\}$, so the $n$th set contains $n$ elements.