Both $\varnothing\subseteq P(\varnothing)$ and $\varnothing\in P(\varnothing)$ true?

Question

Are both $\varnothing\subseteq P(\varnothing)$ and $\varnothing\in P(\varnothing)$ true?
From my understanding $P(\varnothing) = \{\varnothing\}$, so $\varnothing$ is a subset of every set and so $\varnothing\subseteq P(\varnothing)$ is true.
Since $P(\varnothing) = \{\varnothing\}$, then $\varnothing\in P(\varnothing)$ is true too.

Am I right?

Answer 1

Yes.

It is true that $\varnothing\subseteq A$ for any set $A$.

It is also true that $\varnothing \in \mathscr{P}(B)$ for any set $B$.

In particular, your statement follows by taking $A=\mathscr{P}(\varnothing)$ and $B=\varnothing$.

Addendum:

You might be wondering how a set $X$ can satisfy both $X\in\mathscr{P}(X)$ and $X\subseteq\mathscr{P}(X)$. The first is true for any set $X$. So the real question is, for what sets $X$ is it true that $X\subseteq\mathscr{P}(X)$?

Do you think there are any nonempty sets $X$ satisfying $X\subseteq\mathscr{P}(X)$?

Answer 2

Yes. Empty set is a subset of every set. Since $P(\emptyset) = \{ \emptyset \}$, $\emptyset \in P(\emptyset)$.

Answer 3

Yes. The definition of power set is set of all the subsets, which means that every subset of $∅$ is in the power set, hence $∅\in P(∅)$. But the definition of subset is every member of set $A$ is in set $B$, there is no members in $∅$ hence $∅\subseteq P(∅)$