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**dwsmith** Let X be a topological space and let Y be a simply ordered set in the order topology. Let $\displaystyle f,g:X\to Y$ be two continuous functions.

Show that the set $\displaystyle \{x\in X:f(x)\leq g(x)\}$ is closed in X.

Since f is continuous, for every open set $\displaystyle V\subset Y$, $\displaystyle f^{-1}(V)=\{x\in X: f(x)\in V\}$ is an open subset of X. Similar, for every open set $\displaystyle U\subset Y$, $\displaystyle g^{-1}(U)=\{x\in X: g(x)\in U\}$.

How do I go from this to showing what needs to be shown?