Given $\mathbb R^{(0,1)} = \{f:(0,1)\to\ \mathbb R \mid f\text{ is a function}\}$ we define the relation $$f \le g \iff f(x) \le g(x), \forall x \in (0,1).$$
The problem requires to show that this is an order relation, which I did, but then it asks if there are any incomparable elements in $\mathbb (R^{(0,1)},\; \le)$ and also it asks if $\mathbb (R^{(0,1)},\; \le)$ is a lattice. I don't really know how to approach this part.
I'm new to order theory, but here's what I tried:
I know that two elements are comparable if $x \le y$ or $y \le x$ which should work here, but then I thought what if some $\lim f(x) = \infty$ for some $x \in (0,1)$, is $f(x)$ still comparable with any function from $\mathbb (R^{(0,1)})?\ (1)$ I can't figure this (1) out.
For the second part where it asks if $\mathbb (R^{(0,1)}, \le)$ is a lattice if the condition from (1) is acomplished then $\mathbb (R^{(0,1)}, \le)$ is a chain (sequence) then we know that every chain (sequence) is a lattice.
Well, did you consider something simple like $f(x)=\frac12$ for all $x\in (0,1)$ and $g(x)=x$ for all $x\in(0,1)$?
As for finding meets and joins of elements, did you consider the obvious candidates $x\mapsto\min(f(x),g(x))$ and $x\mapsto\max(f(x),g(x))$?
You were on a reasonable track with "two elements $f$ and $g$ are not comparable if $f\nleq g$ and $g\nleq f$." By definition then, there would have to exist an $x$ such that $f(x)\nleq g(x)$ and also an $x'$ such that $g(x')\nleq f(x')$. In words: both of the functions have to climb over each other at some point. Hence, the simple example above.