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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

  • 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.

Reflexivity

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

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

Anti-symmetry

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 and $c^2d^2

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

Transitivity

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

$(a,b) \text { } \Sigma \text { } (c,d)$ 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 then $a^2b^2

Conclusion

$(\mathbb Z \times \mathbb Z,\Sigma)$ 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?

  • 0
    @joriki Please $c$onsider $c$onverting $y$our comment into an answer, so that this question gets removed $f$rom the [unanswered tab](http://meta.math.sta$c$kex$c$hange.$c$om/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-25

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