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Let $T$ be the partial order relation defined on $\mathbb{N}\times \mathbb{N}$ by $(a,b)\, T\, (c,d)$ if and only if $a$ is smaller than or equal to $c$ and $b$ is smaller than or equal to $d$. Is it a total order relation?

I'm thinking it's yes but I don't know how to justify?

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closed as off-topic by Namaste, Arnaldo, Shailesh, Daniel W. Farlow, Xam Apr 26 '17 at 3:27

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  • $\begingroup$ Well, what are the factors that lead you to think yes? $\endgroup$ – Namaste Apr 25 '17 at 22:39
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    $\begingroup$ There must be some reason you're thinking yes, no? If you don't have any thoughts that lead you to think yes, then it seems you simply guess. So, how about you start with adding, to your post, how T satisfies a partial order relation, as well as the criteria/definition of a total order. $\endgroup$ – Namaste Apr 25 '17 at 22:44
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It is not a total order relation because you cannot compare $(1,2)$ and $(2,1)$.

This means that you cannot say $(1,2)\, T\, (2,1)$ or $(2,1)\, T \,(1,2)$.

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Take out a piece of paper, and draw the first few points of $\Bbb N\times \Bbb N$ in a square, say up to $5\times 5$ or $6\times 6$. Now colour every point $(a,b)$ such that either $(a,b)T(3,3)$, or $(3,3)T(a,b)$. Can you see that some points are uncoloured, i.e. incomparable to $(3,3)$? What is the conclusion?

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  • $\begingroup$ So it's a no cos a has to be smaller or equal to c and b has to be smaller than or equal to d? $\endgroup$ – Sook Lim Apr 25 '17 at 22:43
  • $\begingroup$ So can you, for instance, compare $(4,2)$ to $(3,3)$ using $T$? $\endgroup$ – Arthur Apr 25 '17 at 22:46

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