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In $\mathbb{R^2}$ we consider the relation

\begin{equation} (x,y)R(a,b)\Longleftrightarrow (x=a \,\text{and} \, y=b)\, \text{or}\,(x^2+y^2<a^2+b^2) \end{equation}

$R$ is an order relation if it is:

1) Reflexive: $\forall (x,y)\in \mathbb{R^2}, (x,y)R(x,y)$:

It's true because $x=x$ and $y=y$

2) Antisymmetric $(x,y)R(a,b) $ and $ (a,b)R(x,y)\Rightarrow \, (x,y)=(a,b)$

It's true because if $x^2+y^2<a^2+b^2$ and $a^2+b^2<x^2+y^2$ then $x=a$ and $y=b$

3) Transitive $(x,y)R(a,b) $ and $(a,b)R(c,d) $

It's true because: $x^2+y^2<a^2+b^2 $ and $a^2+b^2<c^2+d^2$ then $x^2+y^2<c^2+d^2$

So $R$ is an order relation.

Is correct my work?

B. David
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    Your reasoning in 2) is false. For $x = a$ and $y = b$ we don't have $x^2 + y^2 < a^2 + b^2$. – Stefan Mesken Oct 12 '17 at 18:39
  • @JMoravitz The order isn't supposed to be strict - see the definition. – Stefan Mesken Oct 12 '17 at 18:40
  • I missed seeing the "$(x=a~\text{and}~y=b)$ or" part. That being said, this is missing from OP's attempts at proof for $2$ and $3$ as well. – JMoravitz Oct 12 '17 at 18:43
  • @StefanMesken so how can I prove the antisymmetric? – B. David Oct 12 '17 at 18:54
  • You didn't start with the correct order. $(x,y)R(a,b)$ does not mean that $x^2 + y^2 < a^2 + b^2$. There is another possibility included in the definition of $R$ and that one leads to antisymmetrie. – Stefan Mesken Oct 12 '17 at 18:56
  • @StefanMesken By definition $x=a $ and $b=y$ then $(x,y)=(a,b)$ ?? – B. David Oct 12 '17 at 18:56
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    No, that's not the correct order either. You have $(a,b)R(x,y)$ and thus [$a=x$ and $b=y$] or [$a^2+b^2 < x^2 + y^2$]. Similarly for $(x,y)R(a,b)$. Your proof already shows that the second part of this disjunction can't hold. Thus we may conclude that...? – Stefan Mesken Oct 12 '17 at 18:59

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