I understand why this is the case comparatively , if a set is looked at as being a 'box', then the empty set is considered to be an empty box. So the set $\{1,2,3\}=A$ is considered to be a 'box' containing elements $1,2,3$, and $\{1,2,3,\emptyset \}=B$ is the set containing the elements $1,2,3$ and an 'empty box'?
What I struggle with is how this intuition would make sense given the empty set is a subset of every other set.
Also, I know that $\{\emptyset \} \neq \emptyset$, so would it be the case that for the given set $A$, $\emptyset \subset A$, but that $\emptyset \notin A$, as the set containing no elements is not in the set $A$, but $\emptyset \in B$ is true, right?
In the midst of forming this question I thought of a way of looking at this question, but I am not sure whether it is correct:
So continuing with the comparison of looking at a set being a 'box', $\emptyset \subset A$ simply means that the the set $A$ has some 'free space' in the box, and so the set could be visualised by $A=\{1,2,3,\qquad\}$?, and every set\'box' has some 'empty space', whereby the 'empty box' represents simply the collection of that void.
Does this analogy capture the true picture of Why $\{1,2,3\} \neq \{1,2,3,\emptyset \}$ and $\emptyset \subset A$?
Just a very quick question,
Is the power set of $C=\{1,\emptyset\}$,
$$\mathcal{P}(C)=\{\{1\},\{ \emptyset \},\{1, \emptyset \},\emptyset \}?$$