Power set of $\{\{\varnothing\}\}$

Question

$$\mathcal{P}(x)=\{y\mid y\subseteq x\}$$

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

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

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

For $\mathcal{P}(\{\{\varnothing\}\})$ we have: $$\varnothing\subseteq\{\{\varnothing\}\}$$

$$\{\varnothing\}\subseteq\{\{\varnothing\}\}\text{ and}$$

$$\{\{\varnothing\}\}\subseteq\{\{\varnothing\}\}$$

Therefore $\mathcal{P}(\{\{\varnothing\}\})=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\}$

Is this answer correct? Unsure with this topic, just need some verification, thanks.

Answer 1

This one $$\{\varnothing\}\subseteq\{\{\varnothing\}\}$$ is incorrect.
$A\subseteq B$ means that for every $x\in A$ it is true that $x\in B$.
$\{\varnothing\}$ contains one element: $\varnothing$, but $\{\{\varnothing\}\}$ doen not contain $\varnothing$ so $$\{\varnothing\}\not\subseteq\{\{\varnothing\}\}$$

The other two are correct. $$\mathcal{P}(\{\{\varnothing\}\})=\{\varnothing,\{\{\varnothing\}\}\}$$

And as always: power set of set which contains one element contains two elements.

Answer 2

What is the power set of a one-element set? Work out $\mathcal{P}(\{a\})$ for the singleton set and substitute $a = \{\varnothing\}$

The fact that $\{\varnothing\}$ is your set element (and is itself a set) is what appears to be confusing you. Forget what it is, it's just an element.