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 $\{\emptyset\}$".

Solution 1:

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.

Solution 2:

No, $\emptyset \in \{\emptyset\}$ but $\emptyset \notin \{\{\emptyset\}\}$.

Solution 3:

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 $\emptyset$.