Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Do you know what is the basis for the order topology? –  Asaf Karagila Feb 25 '11 at 8:18

2 Answers 2

up vote 7 down vote accepted

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?

share|improve this answer
1  
Thanks. It is very nice. –  M.Subramani Feb 26 '11 at 17:20

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.