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Let $T = (\mathbb Z\times\mathbb Z, \Sigma) $ be defined as follows:

$$\begin{aligned} (a,b) \text{ } \Sigma \text { } (c,d) \Leftrightarrow (a,b) = (c,d) \text{ or } a^2b^2<c^2d^2\end{aligned}$$

  • check $\Sigma $ is a partial order and it's not total;
  • search for $\min(T)$, $\max(T)$, minimal and maximal elements, if there're any.

Take $T' = \{(1,6), (-1,1), (0,1), (-1,-1), (2,-3)\} \subset \mathbb Z\times\mathbb Z$ then:

  • draw an Hasse diagram for $(T', \Sigma)$, search for $\min(T')$, $\max(T')$, minimal and maximal elements, supremum and infimum, upper and lower bounds;
  • tell if $(T', \Sigma)$ a totally ordered set;
  • tell if it is a lattice.

In order to prove $(\mathbb Z \times \mathbb Z,\Sigma)$ a partial order set, $\Sigma$ reflexivity, anti-symmetry and transitivity has to be shown.


$\forall (a,b) \in \mathbb Z\times \mathbb Z$

$(a,b) \text { } \Sigma \text { } (a,b) $ as $(a,b) = (a,b)$.


Let $(a,b),(c,d) \in \mathbb Z \times \mathbb Z$, then $\Sigma$ is anti-symmetric if

$(a,b) \text { } \Sigma \text{ }(c,d)$ and $(c,d) \text { } \Sigma \text{ }(a,b) \Rightarrow (a,b) = (c,d)$

hence $\forall (a,b),(c,d)$ should be valid the following

$a^2b^2<c^2d^2$ and $c^2d^2<a^2b^2 \Rightarrow (a,b) = (c,d)$

which clearly can't be, so anti-symmetry is only valid if $(a,b) = (c,d)$.


$\forall (a,b),(c,d),(e,f) \in \mathbb Z \times \mathbb Z$

$(a,b) \text { } \Sigma \text { } (c,d)$ and $(c,d) \text { } \Sigma \text { } (e,f) \Rightarrow (a,b) \text { } \Sigma \text { } (e,f)$

which is true, because (assuming all pairs being distinct):

$a^2b^2 < c^2d^2$ and $c^2d^2<e^2f^2$ then $a^2b^2<e^2f^2$


$(\mathbb Z \times \mathbb Z,\Sigma)$ is actually a partial order set and not total. In $(\mathbb Z \times \mathbb Z)$ we can find all $(a,b) : a = 0 \text { or } b = 0$ are minimal elements. However no maximum, minimum or massimal elements have been found. The Hasse diagram for the above mentioned set is:

Hasse Diagram

$(T', \Sigma)$ is not an ordered set, because $\forall (a,b),(c,d) \in T'$ it has to happen $(a,b) \text { } \Sigma \text{ } (c,d)$ or $(c,d) \text { } \Sigma \text{ } (a,b)$, but for the elements $(-1,1),(-1,-1)$ the $\Sigma$ relation doesn't apply.

  • $\sup(T') = (1,6)$, whereas $(1,6)$ is also a maximal.
  • $\inf(T') = (0,1)$, whereas $(0,1)$ is also a minimal.
  • upper bounds are $\{(a,b) \in \mathbb Z \times \mathbb Z : a^2b^2 > 36\}$.
  • no maximum or minumum or lower bounds found.

My question would be: does everything hold? How about the anti-symmetry that is valid only if two pairs are equal? Does it mean in $T'$ the relation $\Sigma$ desn't really apply? How do I prove if this is a lattice?

share|cite|improve this question
You're replacing $\Sigma$ by only the $\lt$ condition, but it's defined as a disjunction of two conditions. Also, "which clearly can't be, so anti-symmetry is only valid if $(a,b)=(c,d)$" seems to indicate a misunderstanding of what's to be shown here -- you need to show that for all $a,b,c,d$, the left-hand side of the implication (including both conditions, not just the $\lt$ one) implies the equality on the right-hand side. – joriki Jul 16 '12 at 16:02
@joriki I am so sorry but I don't know if I have fully understood what you mean, maybe are you basically suggesting I have explicitly examine the case $(a,b) = (c,d)$, that I did omit because I thought it was trivial? – haunted85 Jul 16 '12 at 16:12
Then I'd announce that at the beginning, especially since everything else is very detailed, so one might not expect that you're leaving things out that you find trivial. Also the point about the sentence "which clearly can't be, so anti-symmetry is only valid if $(a,b)=(c,d)$" remains. It makes no sense to say that antisymmetry is only valid in a certain case, because the antisymmetry contains a universal quantifier (though you didn't write it as such, which you did for the other two conditions; perhaps it would help to clear things if you do them all in the same style). – joriki Jul 16 '12 at 16:20
@joriki Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 25 '13 at 11:38

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