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!

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  • $\begingroup$ This looks fine, and you're correct that it follows immediately from the assumption that the relation \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
  • $\begingroup$ You didn´t need the "consistency of the two orders" either or even that they are orders. Metrizability, connectedness and order topology on $X$ are also irrelevant. $\endgroup$ – Ramiro de la Vega Feb 14 at 13:21

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