# Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$

Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous. Show that the set $A = \{x:f(x)\leq g(x)\}$ is closed in $X$.

I am completely stumped on this problem. As far as I can tell I've either got the task of proving $A$ gathers its limit points, or showing that there is a closed set $V \subset Y$ such that $f^{-1}(V)$ or $g^{-1}(V)$ is equal to A, which would prove $A$ closed since both $f$ and $g$ are continuous.

The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set $V$. The fact that I don't know if $f$ or $g$ are injective means I keep running into the problem of having $f^{-1}(V)$ or $g^{-1}(V)$ give me extra points not in $A$. And even if I were able to construct $V$ this wouldn't guarantee it was closed.

I suspect I may need to use the continuity of both $f$ and $g$ together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks.

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HINT: Let $h:X\to Y:x\mapsto\max\{f(x),g(x)\}$.

1. Show that $h$ is continuous.
2. Note that $A=\{x\in X:h(x)=g(x)\}$.
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This is part (b) of the problem almost word for word, my problem is part (a), I'm not sure what to make of that.. Except that they use min instead of max. –  Tyroshi pleasure barge Jul 1 '12 at 20:42
@Nollie Tré: This is the simplest way I know to prove that $A$ is closed. In connection with (2) of the hint, you may want to recall (or prove) that if $f,g:X\to Y$ are continuous, then $\{x\in X:f(x)\ne g(x)\}$ is open in $X$; this is true for any $X$ and any Hausdorff $Y$. –  Brian M. Scott Jul 1 '12 at 20:53
I certainly believe you, I just wonder if maybe there is an error in the book and parts (a) and (b) were switched. Anyways thanks for the help, I should hopefully be able to figure it out from here. –  Tyroshi pleasure barge Jul 1 '12 at 20:57
@Nollie Tré: That would be my guess (that they’ve been interchanged). And the same basic argument could be used with either $\max$ or $\min$. –  Brian M. Scott Jul 1 '12 at 20:58
Ok after going over this all yesterday I'm pretty sure my book doesn't have (a) and (b) switched. To prove your hint requires the pasting lemma which requires that my set $A$ be closed. So as far as I can tell part (a) is necessary for part (b) and not the other way around. –  Tyroshi pleasure barge Jul 2 '12 at 17:36

To show the set $A$ is closed, we only need to proof

$A^c=\{x\in X:f(x)>g(x)\}$ is open.

Proof: For any point $x \in A^c$, $f(x)>g(x)$; and the order topological space is Hausdorff, then there exist disjoint open sets $U_1$ with $f(x) \in U_1$ and $U_2$ with $g(x) \in U_2$ in $X$, such that for any point $y \in U_1$ and $z \in U_2$, $f(y)>g(z)$. Let $U= f^{-1}(U_1) \cap g^{-1}(U_2)$ is an nonempty open set in $X$: for $x \in U$. Obviously, we see $x \in U \subset A^c$, which implies the set $A^c$ is open.

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On line two of your proof I think you meant "in $Y$" not "in $X$". Also could you explain how it follows that in these two disjoint open sets in $Y$ we always have $f(y) > g(z)$ for any $y\in U_1$ and $z\in U_2$? I don't see how this follows from what you have above it. Thanks. –  Tyroshi pleasure barge Jul 2 '12 at 17:33
It is from the special structer of the space $Y$ with ordinal topology, and I think it's not a difficult thing, therefore, I omitt the process of the proof:) –  Paul Jul 3 '12 at 0:18

Hint for the problem:

1. Recall that $Y$ under the order topology is Hausdorff (Exercise).

2. Show that the complement of $A$ is open. Do this by supposing that $x \notin A$. Then $g(x) < f(x)$. If this is the case, then either there exists $y$ such that $g(x) < y < f(x)$, or (exclusively) there does not exist any $y \in Y$ in between $f(x)$ and $g(x)$.

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