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Let us consider a set $A = \{1, 2, 3, 4, 5\}$ and an element say $1$.

Can we say that $1$ is subset of set $A$.

If not, please explain.


marked as duplicate by M. Winter, Lord Shark the Unknown, Namaste elementary-set-theory Jul 27 '18 at 0:09

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  • $\begingroup$ A set of which every element is a subset is called transitive. Unsurprisingly for something we bothered naming, transitivity is a property only of some sets. If you want to construe natural numbers as subsets of natural transitive sets, look up how we model naturals as ordinals. $\endgroup$ – J.G. Jul 24 '18 at 12:01
  • 2
    $\begingroup$ Formally it depends on how $1$ is defined. If e.g. $1:=\{2\}$ (a far-fetched choice) then element $1$ is also a subset of $A$. $\endgroup$ – drhab Jul 24 '18 at 13:21
  • $\begingroup$ ...and $\{1\}$ is both an element and a subset of $\{1,\{1\},\dots\}$ $\endgroup$ – AccidentalFourierTransform Jul 24 '18 at 13:38

No, we say that

  • $1$ is an element of $A$, that is $1\in A$


  • $\{1\}$ is a subset of $A$, that is $\{1\}\subseteq A$

$1$ can not be subset of $A$ because $1$ is not a set. We can say that a set $B$ is a subset of a set $A$ if every element of B is also element of A. For example, the sets: $\{1, 2\}, \{2, 4, 5\}$ are subsets of A.


Unless you explicitly define another set which contains the element 1, then we cannot say that 1 is the subset of A.

There are exceptions, however.

Let's consider an rather unlikely, but still very much possible, scenario in which we define the symbol '1' as a set with elements {2,5}. In this case, then 1 is definitely a subset of A, because 1 now a symbol which denotes a set that contains the elements which also belongs to A (2 and 5).


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