Let $(X,\succsim)$ be a metrizable and connected, totally ordered topological space with the order topology. Let $\succeq$ be another order relation on $X \times X$ such that the two orders are consistent: $x \succsim y \iff (x,z) \succeq (y,z)$ for all $z$. Moreover we assume continuity of $\succeq$, that is 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.

If $(x_n)_n$, $(y_n)_n$ converge monotonically to $x$ and $y$ respectively, and $(w_n)_n$, $(z_n)_n$ converge monotonically to $w$ and $z$ 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 looks quite trivial by definition and my proof is the following:

Denote the set $A:=$ {$(x,y,w,z) \in X \times X \times X \times X : (x,y) \succeq (w,z)$} and pick a sequence of points in it (note that as long as $X$ is metrizable, $A$ preserves all the "good" properties of $X$), that is pick $(x_n,y_n, w_n,z_n)_n \in A^{\mathbb{N}}$ converging monotonically to $(x,y,w,z)$, i.e. $(x_n)_n \rightarrow x$, $(y_n)_n \rightarrow y$, $(w_n)_n \rightarrow w$, $(z_n)_n \rightarrow z$. Then I want to show that $(x,y,w,z) \in A$. As long as $A$ is closed (by assumption) the conclusion follows. Am I missing something? It seems too easy. Moreover, I it seems to me I do not even need the "monotonic" convergence of the sequence. Thanks!

`\succeq`

is closed. That's all you need to "pass to the limit in an inequality". The monotonicity is not important. $\endgroup$ – Nate Eldredge Feb 14 at 13:20