Is $1$ a subset of $\{1\}$ 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...
 A: As @AsafKaragila already said, you can define natural numbers as sets. In fact, in most axiomatic set theories, that's the only way of defining numbers, since every element of a set is a set.
As for your specific question, suppose that $1\subset\{1\}$. This leaves us with two options:


*

*$1=\emptyset$ and the inclusion trivially holds.

*$1\neq\emptyset$. Since $\{1\}$ has only two subsets, $1=\{1\}$.
Option 1 is possible, since identifying $1$ with the empty set is perfectly valid, but it's much more natural to identify $0$ with the empty set.
Option 2 just "looks wrong", but in elementary set theory, we can't really make any statements beyond that. In any axiomatic set theory that includes the Axiom of Regularity, $1=\{1\}$ is false since $x\notin x$ for every set $x$.
A: $1 \in \{1\}$
$\{1\} \subseteq \{1\}$
As you said, 1 is an element of $\{1\}$, not a set.
A: On both the cases, 1 is element of {1} & {1} is element of {{1}}.
                   i.e.  1 ∈{1} & 
                        {1}∈{{1}}
And, in case of subset, {1} is subset of {1}. 
                   i.e. {1}⊆{1}
A: 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 $1 \in \{1\}$.
A: $1$ is an element of $\{1\}$, $\{1\}$ is a subset of $\{1\}$, $\{1\}$ is an element of $\{\{1\}\}$ and $\{\{1\}\}$ is a subset of $\{\{1\}\}$. 
A: $1$ is generally not a subset of $\{1\}$, since $1$ is a natural number (or a real number, or whatever) and not a set. These objects are of two different types.
 
But there is something to be said here. We can represent numbers using sets. We can declare that $0$ is $\varnothing$, and that $1=\{0\}$ or $\{\varnothing\}$, and that $2=\{0,1\}$ and so on. Then a number is a set.
Still that doesn't mean that $1\subseteq\{1\}$. This would very much depend on the representation of $1$ as a set.
So while "working mathematics" is typed (meaning the type of objects matters), we can also work in an untyped environment, where everything has the same type (for example, everything is a set).
A: It is ambiguous but not uncommon to see a set with a single member referred to as simply that member, in contexts where the author would expect the reader to understand that only sets are under discussion.
