How to arrange the following sets? Given the set $\mathcal{P}(\mathbb{N})$ for the following order: For $A,B \in \mathcal{P}(\mathbb{N}) $ applies 
$A \leq_{set} B \Longleftrightarrow_{def} A = B$ or$ ~$ min$(A\triangle B)\in B$ 
We presuppose the natural ordering of $\mathbb{N}$ 
Now I have to order the following sets ascending in terms of $\leq_{set}$
$\emptyset$,$~~ \mathbb{N}$,$ ~~ \{ ~ n ~ | ~ (\exists m \in \mathbb{N})[n=2m] \} $,$~~\{0,1\}$, $~~ \mathbb{N}$,$ ~~ \{ ~ n ~ | ~ (\exists m \in \mathbb{N})[n=m^2] \} $, $~~ \{0\}$
I thought the following order is correct:
$\emptyset \leq_{set} \{0\} \leq_{set} \{0,1\} \leq_{set} ~~ \mathbb{N} \leq_{set}  ~~ \{ ~ n ~ | ~ (\exists m \in \mathbb{N})[n=2m] \} $,$~~\{0,1\} \leq_{set}  ~~ \{ ~ n ~ | ~ (\exists m \in \mathbb{N})[n=m^2] \} $ 
but I think I didn't really understand the condition min$(A\triangle B)"\in B"$. 
Hope somebody can help
 A: For all $A,B \in \mathcal{P}(\Bbb{N})$, you have that $A\triangle B \subseteq \Bbb{N}$. If this is empty, then $A=B$, otherwise it has a minimum element.
But now, this minimum belongs either to $A$ or to $B$ (not to both!). If it belongs to $A$, then you say that $A$ is bigger than $B$ ($B \le_{set} A$), otherwise $B$ is bigger than $A$.
I don't check that $\le_{set}$ is actually an order relation, but from what it is written above, you can easily see that $\le_{set}$ is a total order. Let's see some examples.


*

*$\emptyset$ is the minimum element. This is because for all $A$ you have $$A \triangle \emptyset = A$$ and $\min A \in A$: hence $\emptyset \le_{set} A$.

*$\Bbb{N}$ is the maximum element. This is because for all $A$ you have $$A \triangle \Bbb{N} = A^c$$, and $\min A^c \in \Bbb{N}$.

*In general, if $A \subset B$, then $A \le_{set} B$: this is because
$$\min (A \triangle B) = \min (B \setminus A) \in B$$
Now, denote by $S= \{ \mbox{squares} \}$ and $E= \{ \mbox{even numbers} \}$. Which is the biggest between the two? Well,
$$\min (S \triangle E) = 1 \in S$$
so that $E \le_{set} S$.
Proceeding like this, and recalling 3., you can order
$$\emptyset \le_{set} \{ 0\} \le_{set} E \le_{set} \{0,1 \} \le_{set} S \le_{set} \Bbb{N}$$
