reflexive, symmetric, and transitive relations proof Let $A = \{1, 2, 3, ... , n\}$ where $n$ is a positive integer. Let $F$ be the set of all functions from $A$ to $A$. Let $R$ be the relation on $F$ defined by:
for all $g, f \in F, fRg$ if and only if $f(i) \lt= g(i)$ for some $i \in A.$ Let $I$A : $A \to A$ be the identity function on $A$ defined by $I$A$(x)$ = $x$ for all $x \in A$.
a) Is $R$ reflexive? symmetric? transitive? Prove your answers. 
b) How many elements $f \in F$ are there so that $I$A$Rf$? Explain. 
c) How many elements $f \in F$ are there so that $fRI$A? Explain.
d) How many elements of $f \in F$ are there so that $fRI$A and $f$ is onto? Explain.
My attempt to answer:
a)
 We know a relation $R$ is reflexive iff for all $x\in A$, $xRx$
 We know a relation $R$ is symmetric iff for all $x1, x2\in A$, if $x1Rx2$ then $x2Rx1$
 We know a relation $R$ is transitive iff for all $x, y, z\in A$, if $xRy$ and $yRz$ then $xRz$
Now, using these definitions, we should be able to answer a). 
$R$ is reflexive: 
Let $f\in F$. 
Then $f(1)\leq f(1)$, so $fRf$.

If $n=1$ then $F$ contains exactly one element and it is obvious that in that case $R$ is symmetric and transitive. This as a consequence of the fact that it is reflexive. So let us assume that $n>1$ from here.

$R$ is not symmetric: 
Let $f\in F$ be prescribed by $i\mapsto1$ and $g\in F$ by $i\mapsto2$. 
Then $fRg$ but not $gRf$.

$R$ is not transitive: 
Let $f,g,h\in F$ with $f(1)=f(2)=2$, $g(1)=2\wedge g(2)=1$ and $h(1)=h(2)=1$. 
Then $fRg\wedge gRh$ but not $fRh$.
Now b)
We know that Identity function returns the same values as its parameter, right? So, if you pass f(1) = 2, then identity function of f IA (2) = 1
so the question is how many elements $f \in F$ are there so that $fRI$A? Since, the IA is defined as:
IA(x) = x, this means there are n number of elements?
 A: First of, reflexivity:
You wrote down the definition of reflexive, which is your starting point. Now, it's time to actually do some work.
You want to prove or disprove the statement:

For all $f\in F: fRf$, (because the relation is defined on $F$)



*

*If you want to prove the statement, then by definition, you therefore want to show that for every function $f$, there exists some $i\in A$ for which $f(i)\leq f(i)$. 

*If you want to disprove the statement, then you must prove the negation of the statement. The negation of the statement 
$$\forall f\in F: \exists i\in A: f(i)\leq f(i)$$
is the statement $$\exists f\in F:\forall i\in A: f(i)>f(i)$$


Now, do you want to prove or disprove the statement?

Now, on to symetric. If you want to prove the statement or disprove it? Do you think the relation is or is not symmetric?
A: Okay, here is the answer to my own question: 
$a) $
$R$ is reflexive: 
Let $f\in F$. 
Then $f(1)\leq f(1)$, so $fRf$.

If $n=1$ then $F$ contains exactly one element and it is obvious that in that case $R$ is symmetric and transitive. This as a consequence of the fact that it is reflexive. So let us assume that $n>1$ from here.

$R$ is not symmetric: 
Let $f\in F$ be prescribed by $i\mapsto1$ and $g\in F$ by $i\mapsto2$. 
Then $fRg$ but not $gRf$.

$R$ is not transitive: 
Let $f,g,h\in F$ with $f(1)=f(2)=2$, $g(1)=2\wedge g(2)=1$ and $h(1)=h(2)=1$. 
Then $fRg\wedge gRh$ but not $fRh$.
b) $n^n$
c) $n^n$
d) $n!$
A: Hint: Get your hands dirty by writing down some actual examples of functions; don't try to prove things just yet (we're just working on some intuition for now). For example, consider the following functions $f,g,h \in F$, where for concreteness I fix $n = 4$. Is $fRg$? Is $gRh$? Is $fRh$? Is $hRf$?
$$
\begin{array}{c|c|c|c}
i & f(i) & g(i) & h(i) \\ \hline
1 & 2 & 3 & 1 \\
2 & 4 & 1 & 3 \\
3 & 4 & 3 & 1 \\
4 & 4 & 3 & 1
\end{array}
$$
