Let $X$ be an arbitrary topological space, let $Y$ be an ordered set in the order topology, and let the maps $f, g \colon X \to Y$ be continuous. Then how to show that the set $S$ given by $$S \colon= \{ \ x \in X \ \colon f(x) \leq g(x) \ \}$$ is closed in $X$?
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I'd like the proof to be as elementary and as direct as possible. – Saaqib Mahmood Mar 24 '15 at 06:21
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@Brian M. Scott how were you able to spot so quickly that this question of mine is a duplicate? I too would like to be able to do this so as to avoid such duplications in the future. – Saaqib Mahmood Mar 24 '15 at 06:56
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@Saaqib: I actually didn’t spot it: someone else, who has now deleted the comment, did, and I checked and saw that he was right. The earlier one was from before I went away for almost a year, and I’d completely forgotten that it existed! – Brian M. Scott Mar 24 '15 at 07:08