Prove that this function is a order relation Let M be a set and denote by $\mathbb{F}$ the set of all functions $f:M \rightarrow \mathbb{R}$, show that,
$ f \leq g \Leftrightarrow \forall a \in M : f(a) \leq g(a) $
is a order relation for $(f,g) \in \mathbb{F} \times \mathbb{F}$
Is this a total order?
So I have to show
For reflexivity 
1) $\forall a \in M f(a) \leq f(a)$ (it is clear that this holds)
Transitivity
2) I need to show that if $f \leq g$ and $g \leq h$ then $f \leq h$. It is clear that this holds, but I have difficulties proving this formally. 
Antisymmetric
3) I need to show that if $f \leq g$ and $g \leq f$ then $f=g$ (Here I also do not know how to show this)
For the total order I need to show that 
4) for (f,g) 
$ f \leq g$ OR $g \leq f$
Help and hints would be appreciated
 A: I assume you need to show that $\le_M$ is a partial order on $\mathbb F$. $\le_M$ cannot be a total order as it is not nesseceraly connex. For example, if $a\in M$ and $b\in M$ and $f(a)=0$, $f(b)=1$, $g(a)=1$ and $g(b)=0$, then neither $f\le_M g$ nor $g\le_M f$. Now we will prove it is a partial order.


*

*Reflexivity
$$\forall a\in M (f(a) = f(a))\Rightarrow \forall a\in M (f(a) \le f(a))\Rightarrow f\le_M f$$

*Antisymmetry
$$f\le_M g\wedge g\le_M f\Rightarrow\forall a\in M (f(a) \le g(a))\wedge\forall a\in M (g(a) \le f(a))\Rightarrow\forall a\in M (f(a) \le g(a)\wedge g(a)\le f(a))\Rightarrow\forall a\in M (f(a) = g(a))\Rightarrow f=g$$

*Transitivity
$$f\le_M g\wedge g\le_M h\Rightarrow\forall a\in M (f(a) \le g(a))\wedge\forall a\in M (g(a) \le h(a))\Rightarrow\forall a\in M (f(a) \le g(a)\wedge g(a)\le h(a))\Rightarrow\forall a\in M (f(a) \le h(a))\Rightarrow f\le_M h$$
All of the actions under the qualifiers follow from the rules of real numbers, because $f(x)\in\mathbb R$ for any $f\in\mathbb F$ and $x\in M$.

