# 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

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.
1. Reflexivity $$\forall a\in M (f(a) = f(a))\Rightarrow \forall a\in M (f(a) \le f(a))\Rightarrow f\le_M f$$
2. Antisymmetry $$f\le_M g\wedge g\le_M f\Rightarrow\forall a\in M (f(a) \le g(a))\wedge\forall a\in M (g(a) \le f(a))\Rightarrow\forall a\in M (f(a) \le g(a)\wedge g(a)\le f(a))\Rightarrow\forall a\in M (f(a) = g(a))\Rightarrow f=g$$
3. 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 $$x\in M$$.