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 $f(i) \leq g(i)$ 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 $x \in A$.
Is $R$ reflexive? symmetric? transitive? Prove your answers.
Can someone help answer these please?
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 $x_1, x_2\in A$, if $x_1Rx_2$ then $x_2Rx_1$
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 $fRh$.