Hi there, I have a question relating to partial orders, I can prove only part of the question but can't seem to finish it off, any help would be greatly appreciated!

Let T denote the set of real-valued functions that are differentiable at the point x=0 we define a relation R on T by 'f R g iff $\displaystyle f'(0) \le g'(0)$'. Does R give rise to a partial ordering of T? If not, then which of reflexive, antisymmetry, transitive fails?

So for R to be a partial order on T then R must be reflexive, antisymmetric and transitive.

I can show R is reflexive, for example, let h be any real valued function that is differentiable at the point x=0, ie, $\displaystyle h \in T$

Thus $\displaystyle h'(0) \le h'(0)$ so $\displaystyle (h,h) \in R$ and thus R is reflexive.

I am unsure of how to go about showing R is antisymmetric and transitive (or how to go about showing if one of these fails)

Thanks again!

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Actually after thinking more, I can show R is a partial order but the answer says it's not?

I showed R is antisymmetric like this:

If $\displaystyle (f, g) \in R$ and $\displaystyle (g,f) \in R$ then it is clear that f'(0)=g'(0)

I also showed R is transitive like this:

If $\displaystyle f'(0) \le g'(0)$ and $\displaystyle g'(0) \le z'(0)$ then $\displaystyle f'(0) \le z'(0)$ and thus $\displaystyle (f, z) \in R$

So where in my working have I gone wrong...? :S