Difference between $\emptyset$ and $\{\emptyset \}$ I was tasked to write the following by listing the elements between braces, $$\{X\subseteq\mathscr{P}(\{1,2,3\}): |X|\leq 1\}$$ $\{\emptyset \}$
My textbook gives the solution: $$\big\{\emptyset, \{\emptyset\}, \{\{1\}\}, \{\{2\}\}, \{\{3\}\}, \{\{1, 2\}\}, \{\{1,3\}\}, \{\{2,3\}\}, \{\{1,2,3\}\} \big\}$$
I got everything except the elment $\{\emptyset\}$. What's the difference between $\emptyset$ and $\{\emptyset\}$? Isn't $\emptyset=\{\}$, and likewise $\{\emptyset\}=\{\{\}\}$ which should be equivalent to $\{\}$?
 A: First one is a an empty set and the second one is a set containing only one element - an empty set.
If you want an analogy from the real world - imagine (a) an empty bag and then imagine (b) a bag that has another bag inside, which is empty.
The reason that you have $\emptyset$ as one of the answers is that an empty set is a subset of any set, so it qualifies too.
A: The notation $\emptyset$ is the set with no elements, but it is a set by itself, it does not mean that $\emptyset$ is nothing.
For example: the zero vector $\overset{\to}{0}$ in normed space $(V,\|\cdot\|)$ is the only vector $v$ with $\|v\|=0$ (the norm of a vector is a "scale" of it "size"), $\overset{\to}{0}$ is an element in the normed space, and one can think about $\overset{\to}{0}$ as "nothing", because $\forall v\in V: v+\overset{\to}{0}=v$, but it does not mean that $\overset{\to}{0}$ is nothing.
By the same intuition, the empty set is not nothing, it is just a wallet with no money, which leads to $\{\emptyset\}\neq\emptyset$, because the set $\{\emptyset\}$ is a set with one element, and $\emptyset$ has zero elements.
