1. ## Question about ordered topology

This is a Munkres's problem: Let Y be an ordered set with the ordered topology and f,g functions from X to Y, show that the set$\displaystyle \{x:f(x)\leq g(x)\}$ is closed. I can show this, for intance, when Y=R(real numbers set) but I can't prove this if Y is just an ordered set because it may have consecutive elements. Any comments will be appreciated

2. Originally Posted by facenian
This is a Munkres's problem: Let Y be an ordered set with the ordered topology and f,g functions from X to Y, show that the set$\displaystyle \{x:f(x)\leq g(x)\}$ is closed. I can show this, for intance, when Y=R(real numbers set) but I can't prove this if Y is just an ordered set because it may have consecutive elements. Any comments will be appreciated
Do you want a solution or advice? My advice would be to show that it's compliment is open. It's much more obvious.

3. yes that's what I did but found the problen I mentioned earlier

4. Originally Posted by facenian
yes that's what I did but found the problen I mentioned earlier
Write your argument, I don't (although I haven't admittedly gone through the whole proof) see where you'd want to use that.

5. the proof would be $\displaystyle \{x:f(x)>g(x)\}=\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})$
the problem is that if Y has consecutive elemets then the inclusion
$\displaystyle \{x:f(x)>g(x)\}\supset\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})$ seems to be strict

6. Originally Posted by facenian
the proof would be $\displaystyle \{x:f(x)>g(x)\}=\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})$
the problem is that if Y has consecutive elemets then the inclusion
$\displaystyle \{x:f(x)>g(x)\}\supset\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})$ seems to be strict
Let $\displaystyle W=\{x:f(x)>g(x)\}$. To show W is open in X, I shall show that for every $\displaystyle x_0 \in W$, there exists an open set containing $\displaystyle x_0 \in W$ and is contained in W.

As you said, there are two cases to consider. Let $\displaystyle x_0$ be an arbitrary point in $\displaystyle W$.

1. There is no element between $\displaystyle f(x_0)$ and $\displaystyle g(x_0)$ such that $\displaystyle f(x_0) > g(x_0)$. Then, $\displaystyle f^{-1}(g(x_0), \infty) \cap g^{-1}(-\infty, f(x_0))$ is an open set containing $\displaystyle x_0$ and is contained in W, since f and g are continuous, Y is an ordered set in the order topology, and an intersection of open sets is open.

2. There exists element(s) between $\displaystyle f(x_0)$ and $\displaystyle g(x_0)$ such that $\displaystyle f(x_0) > g(x_0)$. Let z be such an element $\displaystyle f(x_0) > z > g(x_0)$. Then, $\displaystyle f^{-1}(z, \infty) \cap g^{-1}(-\infty, z)$ is an open set containing $\displaystyle x_0$ and is contained in W.

Thus W is open in X and X-W is closed.