# Proof that $(\mathbb{N}\times\mathbb{N},\preceq), (a,b)\preceq(c,d)\iff a\leq c\wedge b\geq d$ is a poset

I'm having a trouble accepting this anti-symmetric property as proof. It seems faulty to me because it's assuming that the relation is anti-symmetric from the outset and not proving that it is.

Anti-symmetric:

If $(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$, $(a,b)\preceq(c,d)\wedge(c,d)\preceq(a,b)$, then $a\leq c$, $b\geq d$, $c\leq a$, $d,\geq b$. Therefore $a=c\wedge b=d$.

Is this fine? I would have thought you'd need to show something along the lines of

• Anti-symmetric: $\forall(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$: \begin{align*} (a,b)\preceq(c,d)&\iff a\leq c\wedge b\geq d\\ &\iff \end{align*}

but now I'm stuck. So maybe the above method is right, but it still doesn't feel right.

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The proof only assumes from the outset that the usual order on $\mathbb N$ is anti-symmetric.

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OK, I see... I can use my method of proving it by changing it a big.. I forgot to add the fact that $(c,d)\preceq(a,b)$. – agent154 Mar 1 '13 at 16:25
@agent154: If you want to add an answer in your own words, you should post a new answer. Don't try to edit it into other people's posts. – Rahul Mar 1 '13 at 16:41
I thought it would be more fair to the poster to get credit for their response – agent154 Mar 1 '13 at 16:50

Note that $\preceq$ is not $\leq$. The former is what you have to show to be anti-symmetric whereas the latter is given as anti-symmetric.

So if $(a,b)\preceq(c,d)\preceq(a,b)$ then $a\leq c\leq a$ and $b\geq d\geq b$. Now what can we deduce?

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After looking at it a bit, it does seem right... and here's a way of showing it in the method I was looking for:

• Anti-symmetric: $\forall(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$: \begin{align*} (a,b)\preceq(c,d)\wedge(c,d)\preceq(a,b)&\Longrightarrow a\leq c\wedge b\geq d\wedge c\leq a\wedge d\geq b\\ &\Longrightarrow a\leq c\wedge c\leq a\wedge b\geq d\wedge d\geq b\\ &\Longrightarrow a=c\wedge b=d \end{align*}
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