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!