How to prove that this relation is antisymmetric? Let $F$ be the set of all functions $f : \mathbb{R} \to \mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ \mathbb{R} $.
Prove that '$c$' is a partial order.
I have proved that the relation is reflexive. Now I must prove that it is antisymmetric (and later transitive). 
My working thus far: 
Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$. 
Now, how to prove that $a = b$? There is no indication that $f$ is an injective function. 
 A: I think you are confuse about your own notation.  You define the relation as

Let $\mathcal{F}(\mathbb{R},\mathbb{R})$ be the set of all functions $f: \mathbb{R}\to \mathbb{R}$. So define $\color{blue}{c}$ as the relation defined on $\mathcal{F}(\mathbb{R},\mathbb{R})$ as, for $f,g \in \mathcal{F}(\mathbb{R},\mathbb{R})$ then $f \color{blue} c g \Leftrightarrow f(x)\leq g(x) \, \forall x \in \mathbb{R}$.

Now take a look in your proof: You say that $a,b \in \mathcal{F}(\mathbb{R},\mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it
Suppose $a,b \in \mathcal{F}(\mathbb{R},\mathbb{R})$. We want to prove that $(a \color{blue }c b $ and $b \color{blue}c a )$ implies $a = b$. Now what you've done is 


*

*$a \color{blue}c b \implies \color{red}\forall x \in \mathbb{R}, a(x)\color{red}\leq b(x)$

*$b \color{blue}ca \implies \color{red}\forall x \in \mathbb{R}, b(x)\color{red}\leq a(x)$
And then $\color{red}{(1)\text{ and } (2)\text{ together}}$ implies that $\forall x \in \mathbb{R}, a(x) = b(x)$ wich implies that $a = b \in \mathcal{F}(\mathbb{R}, \mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.
A: The relation is over the set of functions, $F$, and not on the set of real numbers, $\mathbb R$.
So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $\mathbb R$ to $\mathbb R$.
In your case, $a c b$ means $a \le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) \le f(b)$.
Hence, showing $f=g$ is sufficient.
A: Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f \prec g$ and $g \prec f$ then $f(x) \leq g(x)$ and $f(x) \geq g(x)$ for all $x \in R$. This means $f=g$. No need for injectivity :)
