Partial Orders

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 $f'(0) \le g'(0)$ '. 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, $h \in T$

Thus $h'(0) \le h'(0)$ so $(h,h) \in R$ 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 $(f, g) \in R$ and $(g,f) \in R$ then it is clear that f'(0)=g'(0)

I also showed R is transitive like this:

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

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

Consider the functions $f(x)=x^2~\&~g(x)=cos(x)$ .
Is the relation antisymmetric?
i.e. Does $f=g?$

Why don't you start by writing what it means for any relation R on T to be antisymmetric and then substituting the actual definition for the letter R in that expression? You will get a proposition of the form, "For every two functions f and g in T, ...". Then we can decide whether it is true or false.

Oh right, sigh, what a careless mistake by me, thank you Plato (you're always helping me ^_^) and emakarov.

I assume my reasoning behind reflexivity and transitivity are valid...?

Quote:

I assume my reasoning behind reflexivity and transitivity are valid...?

Yes.