1. Transitve relation help

I need to prove the intermediate value theorem using the following lemma:

Lemma: Let be a transitive relation on the interval . If each has a neighborhood such that whenever and , then .

So, the main goal is to actaully produce a transitive relation $\displaystyle rho$, but I am having no such luck. I'd assume $\displaystyle \leq$ or $\displaystyle <$ ought to work, but I am not sure what other conditions are needed to make the relation work in the proof.

Also, does this prove the lemma given above?

Let be arbitray and be a transitive relation on . By our hypothesis, we can find a neighborhood of such that whenever and . But, if we condsider the set it is clear that exists and because is compact . Since must share a transitive relation with and a transitive relation with , it must be that case that .

Note: I have posted this question in the analysis thread, but I thought it might actaully fit better here, given it really comes down to using the proper relation.

Thanks

2. Interesting use of relations in calculus.

Originally Posted by Danneedshelp
Also, does this prove the lemma given above?

Let be arbitray and be a transitive relation on . By our hypothesis, we can find a neighborhood of such that whenever and .
I don't understand the transition from here to the next statement below. It is strange that you abandon $\displaystyle x$ that you have chosen, and in the following $\displaystyle x$ appears only as a bound variable.
But, if we condsider the set it is clear that exists and because is compact .
To have one has to show that there are points in $\displaystyle A$ that are arbitrarily close to $\displaystyle b$, i.e., that $\displaystyle A$ does not, for example, begin with $\displaystyle a$ end end with $\displaystyle (a+b)/2$. That, I think, is the main thing that one needs to show.

I think this lemma is valid. Since for each point we have a neighborhood, you can choose a finite covering of $\displaystyle [a,b]$. Then, e.g., by induction on the number of neighborhoods in this covering it is easy to show that $\displaystyle a\rho b$.