Partial ordering of functions

Question

Let $X$ be the set of all real-valued functions $x$ on the interval $[0,1]$ and let $x \leq y$ mean that $x(t) \leq y(t)$ for all $t \in [0,1]$. Does it define a partial ordering/ total ordering? Does $X$ have maximal elements?

So, to show that $ \leq $ is a partial ordering on $X$:

i)It's clear that for all $x(t)$ in the interval, $x(t) \leq x(t)$.

ii)$x(t) \leq y(t)$ and $x(t) \geq y(t)$ implies that $x(t) =y(t)$.

iii)If $x(t) \leq y(t)$ and $y(t) \leq z(t)$ then we get $x(t) \leq z(t)$

How can we show that $\leq$ is a total ordering and what are the maximal elements? Can we say that the function $g(t) = 1$ is a maximal element? or is it an upper bound? Edit: I think it's not a total order since we may not able to compare any two elements.

Answer 1

Hint For $\leq$ to be a total order, we must be able to order all real-valued functions on $[0,1]$. But what about

$$x(t) = t$$ $$y(t) = 1-t$$

Can we say that $x \leq y$ or that $y \leq x$? Or neither? (What does that mean?)

Answer 2

We can't compare them as we mentioned above, so that it's not a total order:

Let's consider the functions $x(t)=t$ and $y(t)=|1−t|$. First one is increasing and other one is decreasing, neither $x≮y$ nor $y≮x$.