Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$? $\{\emptyset\}$ is a set containing the empty set. Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?
My hypothesis is yes by looking at the form of "the superset $\{\{\emptyset\}\}$" which contains "the subset $\{\emptyset\}$".
 A: It's easier to just replace everything by variables, then it's clearer.
Set $a=\varnothing, b=\{\varnothing\}=\{a\}$ and $c=\{\{\varnothing\}\}=\{\{a\}\}=\{b\}$.
Now the question is whether or not $\{a\}\subseteq\{b\}$. But since $a\neq b$, it's easy to see that the answer is negative.
A: No, $\emptyset \in \{\emptyset\}$ but $\emptyset \notin \{\{\emptyset\}\}$. 
A: {{∅}} has only one element, namely {∅}, it implies there are only the two trivial subsets, {{∅}} which contains all elements and ∅ which contains none.
formally all subsets of {{∅}}
{{∅}}⊆{{∅}}
∅⊆{{∅}}
A: For such abstract questions, it is important that you stick to the definitions of the involved notions.
By definition, $A$ is a subset of $B$ if every element contained in $A$ is also contained in $B$.
Now we look at $A = \{\emptyset\}$ and $B = \{\{\emptyset\}\}$.
$A$ has exactly one element, namely $\emptyset$.
This is not an element of $B$, since the only element of $B$ is $\{\emptyset\}$ and $\emptyset \neq \{\emptyset\}$ (see below).
So $A$ is not a subset of $B$.
Why $\emptyset \neq \{\emptyset\}$?
By definition, two sets are equal if they contain the same elements. However, $\emptyset$ is the empty set without any element, but $\{\emptyset\}$ is a $1$-element set with the element $\emptyset$.
A: Think of {∅} as a box containing an empty box, then
{{∅}} is a box containing a box which contains an empty box, in which case
{{∅}} is box containing {∅}, which means
{∅} is subset of {{∅}}
