Show with a counterexample that $A\nsubseteq A^A$ Definition
A set $R$ is a (binary) relation if any its element is a ordered pair, that is $z\in R$ if and only if there exist $x$ and $y$ such that $z=(x,y)$. In particular if $R$ is contained in the cartesian product of two set $A$ and $B$ we say that $R$ is a relation of between $A$ and $B$ and if $B$ is equal to $A$ we say that $R$ is a relation in $A$.
Definition
If $R$ is a relation we call domain of $R$ that set whose elements are a first coordinate of some pair of the relation, that is
$$
\text{dom}\,R:=\{x:\exists\,y\,\text{such that}\,(x,y)\in R\}
$$
and analogously the range of $R$ is that set whose elements are the second coordinate of the relation, that is
$$
\text{rank}\,R:=\{y:\exists\,x\,\text{such that}\,(x,y)\in R\}
$$
So it is possible to prove that the above two defined set exist using the $ZFC$'s formalism but this now has not matter so we proceed to give the following well know definition.
Definition
A function $f$ is a relation such that if $(x,y)$ and $(x,z)$ are such that $(x,y),(x,z)\in f$ then $y=z$.
Definition
Let $A$ and $B$ sets. So the set whose element are functions from $A$ to $B$ is denoted by the symbol $B^A$.
Again it is possible to prove that the above defined set exist using $ZFC$'s formalism but now this again has not matter.
So by the previous things I thought that any element $a$ of a (not empty) set $A$ could be regarded as a costant function so that I ask to me if $A$ is a subset of $A^A$ but this seems to me very stranger and so I would like to know the opinon of anyone with more competences of mine. In particular I have observed that if $a\in A^A\cap A$ then $a\subseteq A\times A$ and so either $a=\emptyset$ either there exist $c,d\in A$ such that $a=(c,d)\in\{\{c\},\{c,d\}\}$ but by this I did not deduce any contradiction. So could someone help me, please?
 A: Formally, it is false that $A \subseteq A^A$ for a generic set $A$. Take for instance $A = \{*\}$. Then, $A^A = \{\{(*,*)\}\}$ (i.e. the only function in $A^A$ is the identity on $A$), so $* \in A$ but $* \notin A^A$, thus $A \not\subseteq A^A$. Actually, $A$ and $A^A$ are disjoint in this case.
However, there is a canonical injection of $A$ into $A^A$, the one correctly defined by Wuestenfux in his answer:

Each element $a \in A$ can be identified with the constant function
\begin{align}
f_a \colon& \, A \to A
\\
&\,\,x \mapsto a
\end{align}

In this way, each element $a \in A$ can be thought as an element of $A^A$. Therefore, "morally" (but not formally) $A \subseteq A^A$.

Note that there is an error in your argument. If you assume $a \in A^A \cap A$, then $a \subseteq A \times A$, but possibly there is no $a \in A^A \cap A$ and you are assuming something impossible. Indeed, from $a \in A^A \cap A$ it follows that $a \in A^A$ (and hence $a \subseteq A \times A$) and $a \in A$, but we have just seen that this can be impossible, for instance when $A = \{*\}$ and so $A^A \cap A = \emptyset$.
A: Each element $a\in A$ can be identified with the constant function $f_a:A\rightarrow A:x\mapsto a$. In this way, $a$ can be thought as an element of $ A^A$.
