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I have a question.

Let $f,g$ be continuous functions from $X$ to $Y$, $X$ is a topological space and $Y$ a topological space under ordered topology. Then prove that the set $\{x \in X \ | \ f(x) < g(x)\}$ is open. I want to know that what intrisic property of order makes it possible.

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  • $\begingroup$ Do you know what is the basis for the order topology? $\endgroup$ – Asaf Karagila Feb 25 '11 at 8:18
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Look at the map $(f,g)\colon X\to Y\times Y$. This map is continuous because $f$ and $g$ are.

Your set $\{x \in X \mid f(x) < g(x)\}$ is the preimage under $(f,g)$ of the set $\{(a,b) \in Y\times Y \mid a < b\}$.

Now, the only thing that remains to prove is that $\{(a,b) \in Y\times Y \mid a < b\}\subseteq Y\times Y$ is open.

This follows from the definition of the Order topology on $Y$. Do you need help with that?

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Hey i found this somewhere on the net. See problem 5 in the given link:

http://www.math.uiowa.edu/~jsimon/COURSES/M132Fall07/M132Fall07_Exam1_Solutions.pdf

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