Deciding if subset and element statements involving sets are true or false I'm trying to figure out if these statements are true or false:

(1) {∅} ∈ P(A)
(2) {A} ⊆ A
(3) A ⊆ {A}

This is what I think they are:
(1) false

*

*reasoning: ∅ is a set with no elements, but {∅} is a set with one element (∅). Since ∅ is a subset of every set, ∅ is a subset of A. By definition a power set of a set, in this case A, is a set whose elements are subsets of the set A. So since ∅ is a subset of A, $∅ ∈ P(A)$ is true but not {∅} ∈ P(A)
(2) false

*

*reasoning: A is contained in {A}, but {A} is not contained in A, so A ⊆ {A} is true, but {A} ⊆ A is false.

(3) true

*

*reasoning: see previous explaination
Is what I said correct (both the true/false answer and my reasoning)?
 A: $(1)$ is false: $ \emptyset \subseteq A \implies \emptyset \in \mathcal{P}(A) \implies \{\emptyset\} \subseteq \mathcal{P}(A) \implies \color{green}{  \{\emptyset\} \in \mathcal{P}( \mathcal{P}(A))}.$
$(2)$ is false: $A\subseteq A \implies A \in \mathcal{P}(A) \implies \color{green}{ \{A\} \subseteq \mathcal{P}(A)} \implies  \{A\} \in \mathcal{P}( \mathcal{P}(A))..$
$(3)$ is false: $\color{green}{A \in \{A\}}.$
A: 3 is false.  If it were true, then for all x in A, x = A.
Such a set cannot exist because of the axiom of foundation or regularity.
A: The reasoning for 2 is incorrect, because of the statement $A \subseteq \{A\}$, replace it with $A \in \{A\}$. For understanding, let $A = \{1, 2\}$. It's clear that $\{1, 2\} \not\subseteq \{\{1, 2\}\}$. This also shows that correct answer for 3 is false. You can give this example to prove.
A: The third statement should be false because $A \in \{A\}$ and $A \subseteq \{A\}$ are different statements. Note that the set $\{A\}$ has $2^1=2$ subsets: $\varnothing$ and the set itself, $\{A\}.$ Clearly $A$ is neither of those two subsets.
