# The closed set in the product topology implies the convergence of monotonic sequences

Leave $$(X, succsim)$$ be a topological space totally ordered and connected with the order topology. Leave $$succeq$$ be another order relationship in $$X times X$$ such that the two orders are consistent: $$x succsim y iff (x, z) succeq (y, z)$$ for all $$z$$. We also assume the continuity of $$succeq$$, that's the set {$$(x, y, w, z) in X times X times X times X: (x, y) succeq (w, z)$$} is closed in the product topology.

Yes $$(x_n) _n$$, $$(y_n) _n$$ converge monotonously to $$x$$ Y $$and$$ respectively, and $$(w_n) _n$$, $$(z_n) _n$$ converge monotonously to $$w$$ Y $$z$$ respectively, then I want to show $$(x_n, y_n) succeq (w_n, z_n) implies (x, y) succeq (w, z)$$.

To me, the problem seems quite trivial by definition and my test is as follows:

Denote the set $$A: =$$ {$$(x, y, w, z) in X times X times X times X: (x, y) succeq (w, z)$$} and choose a sequence of points in it (keep in mind that whenever $$X$$ is metrizable $$A$$ retains all the "good" properties of $$X$$), that is to choose $$(x_n, y_n, w_n, z_n) _n in A ^ { mathbb {N}}$$ converging monotonously to $$(x, y, w, z)$$that is to say
$$(x_n) _n rightarrow x$$, $$(y_n) _n rightarrow and$$, $$(w_n) _n rightarrow w$$, $$(z_n) _n rightarrow z$$. So I want to show that $$(x, y, w, z) in A$$. While $$A$$ It is closed (of course) the conclusion follows. Am I missing something? It seems very easy. Besides, it seems to me that I don't even need the "monotonic" convergence of the sequence. Thank you!