Considering $(X,T)$ to be a connected topological space, $f$ being a continuous function from $X \to X$ and the binary relation being irreflexive, transitive, complete and open. $X^2$ has the product topology.
I want to prove that the sets $A = \{ x \in X \mid f(x) > x \}$, $B = \{ x \in X \mid x > f(x) \}$ are open. I have multiple questions.
i) My initial thought was to use the fact that, under a continuous function, pullbacks of open sets are open sets but I’m unable to relate the concept of the binary relation to the normal greater-than sign.
ii) My second thought was to prove that $\{ x \in X \mid f(x) = x \}$ is closed, hence its complement is open. Howeverznonce again I’m not sure how the indifference property of relations connects to the equality sign.
iii) The binary relation is open in $X^2$ iff for all $x, y \in X$, $x>y$ implies there exist open neighborhoods $U(x)$ and $V(y)$ such that all elements of $U(x) >$ all elements of $V(y)$. Could I use this theorem in any way to show that $A$ and $B$ are open?
iv) Does $(X,T)$ being a connected space have any significance?
I’d be grateful for any help! I’m very lost here. I’m sorry if the formatting is incorrect.
