Let Y be an ordered set in the order topology. Let $f,g: X \rightarrow Y$ be two continuous function.
Show that the set {$x | f (x) \leq g (x) $} is closed in $X$. Any help would be appriciated. Thanks.
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You can prove $A' = \{x: f(x) > g(x)\}$ is open, and this is readily open for $A' = (f-g)^{-1}((0,\infty))$ is the pre-image of an open set under the continuous function $h = f-g$
DeepSea
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I'm not sure, is the concept of $0$ and $-$ well defined in general order topology? – May 23 '17 at 05:11
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No it is not. It is given Y is ordered set. Not set of real numbers – user426700 May 23 '17 at 05:14