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R is relation over the set of functions continuous in $[0,1]$ that defined $$fRg \Longleftrightarrow f(x) \leq g(x) \rightarrow x\in [0,1]$$ I know that to prove it I need to show that

  1. if for all $a \in A$(the functions set) implies $(a,a)\in R \rightarrow$ Reflexivity
  2. for all $(a,b) \in R , (b,a) \in R \rightarrow a=b $ i.e. Anti - Symmetry
  3. if for all $(a,b) \in R $ and $(b,c) \in R \rightarrow (a,c)\in R $ Transitivity


for reflexivity: $fRf \longleftrightarrow f(x)\leq f(x)$ $R$ reflexivity
for anti symmetry $fRg ,gRf\longleftrightarrow f(x)\leq g(x) \wedge g(x)\leq f(x) \rightarrow f(x)=g(x) $ $R$ anti symmetry what about transitivity?
Thanks!

Ofir Attia
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  • Please correct the confusing three different uses of $\rightarrow$ ('for all', 'that is', and 'implies'), as I did for your other question http://math.stackexchange.com/q/456531/11994. Thanks! – MarnixKlooster ReinstateMonica Jul 31 '13 at 19:22

1 Answers1

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Proving transitivity (as well as reflexivity and anti-symmetry) really just falls out from the fact that $\leq$ is already transitive (as well as reflexive and anti-symmetric).

Choose any continuous functions $f,g,h$ in $[0,1]$ such that $fRg$ and $gRh$. Then we know that for all $x\in [0,1]$, we have $f(x) \leq g(x)$ and $g(x) \leq h(x)$. Hence, by the transitivity of $\leq$, we know that $f(x) \leq h(x)$ so that $fRh$, as desired.

Adriano
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  • OK, thanks, what about the reflexivity and the anti symmetry I wrote, there is something wrong there? – Ofir Attia Jul 31 '13 at 17:50
  • Nope, nothing wrong. All you really needed to do was write out the definitions like you did; no fancy proof tricks were needed. – Adriano Jul 31 '13 at 17:50
  • @Adriano How did you consider the definition of the relation above? I think there is very confusing point there. He noted the definition as $fRg \Longleftrightarrow f(x) \leq g(x) \rightarrow x\in [0,1]$ not as $fRg \Longleftrightarrow f(x) \leq g(x), \forall x\in [0,1]$. – Mikasa Jul 31 '13 at 18:05
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    @BabakS. It's true that I interpreted the OP's definition to mean your second definition. I assumed that the OP made a typo for that part. If the OP could clarify that part, I'd be willing to change my answer. – Adriano Jul 31 '13 at 18:09
  • I changed, Your answer is understandable – Ofir Attia Jul 31 '13 at 19:36