TITLE: Is $1$ a subset of $\{1\}$
QUESTION [12 upvotes]: Is the number $1$ a subset of the set $\{1\}$ just as $\{1\}$ is a subset of the set $\{\{1\}\}$? I'm a little bit confused because $1$ is an element not a set...

REPLY [3 votes]: No. Every natural number is constructed based on set theory. Because of that every natural number is a set. To be more general, because we build up the whole mathematics from set theory, we can say that in mathematics everything is a set. The construcion is the following.

$0 = \emptyset$
$1 = \{ 0 \} = \{ \emptyset \}$
$2 = \{0, 1 \} = \{ 0, \{0\} \} = \{ \emptyset, \{ \emptyset \} \}$
and so on.

Since $1 = \{ 0 \} = \{ \emptyset \}$ and $\{1\} = \{ \{ 0 \} \} = \{ \{ \emptyset \} \}$ and the subsets of $\{ 1 \}$ are $\wp\left( \{ 1 \}\right) = \wp\left(\{ \{ \emptyset \} \}\right)=\{\emptyset,\{ \{ \emptyset \} \}\}$ we can say that $ \{ \emptyset \} \notin \wp\left(\{1\}\right) $ so $\{ \emptyset \} \not\subseteq \{ \{ \emptyset \} \}$ and that is why $1 \not\subseteq \{1\}$.
On the other hand, because we have an axiom of extensionality and since we have element predicate, the fact that $\{ 1 \}$ contains $1$ exactly means that $1 \in \{1\}$.