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Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?

This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) > g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?


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up vote 5 down vote accepted

The set $$U = \{(y_1, y_2) \in Y \times Y \mid\, y_1 \leq y_2 \,\}$$ is closed in $Y \times Y\,$:

Suppose $(z_1, z_2) \in Y \times Y\,$ is not in it, so that $z_1 > z_2$.

If there is some $z_3 \in Y$ such that $z_1 > z_3 > z_2$, the basic open set $(z_3, \rightarrow) \times (\leftarrow, z_3)$ contains $(z_1, z_2)$ and misses $U$, otherwise $z_1$ and $z_2$ are neighbours and $(z_2, \rightarrow) \times (\leftarrow, z_1)$ has this property. Note that all used subsets in $Y$ are open in the order topology.

Now $f \times g: X \times X \rightarrow Y \times Y$ is continuous (standard in the product topology) when $f$ and $g$ are, and your set is exactly

$$(f \times g)^{-1}[U]$$

and thus closed as the inverse image of a closed set under a continuous function.

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Calling a closed set $U$ is strange :-) But I guess it was $U$ for upper half space... – lhf Sep 21 '11 at 12:02
@lhf: it was for "upper" indeed. – Henno Brandsma Sep 21 '11 at 12:07

The complement of $\{x : f(x) \leq g(x)\}$ is $\{x : f(x) > g(x)\}$. Part of the latter set is $$\bigcup_{y\in Y} \{x : f(x) > y > g(x)\},$$ which is open because each set is the intersection of two open sets. What is missing?

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