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Let $A=\{1,2,3,4,5,6\}$ and let's consider the usual order relation given by $\leq$. My textbook includes an image representing this structure:

enter image description here

Given the definition of lattice, I don't understand how this isn't it. When I am looking for the supremum and the infimum for every pair it is obvious that $\forall (a,b)\in A$

$$\begin{aligned} \;a\leq6 \text{ and } b\leq6, 1\leq a \text{ and } 1\leq b\end{aligned}$$

so what is that I don't take into account?

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I think 2,3 don't have a join, and 4,5 don't have a meet? –  sperners lemma Nov 13 '12 at 18:13
    
@spernerslemma I'm trying to figure out this thing myself... and it's kind of confusing. I don't know why they aren't joined! –  haunted85 Nov 13 '12 at 18:15
    
Sperners observation is correct. Both $4$ and $5$ are upper bounds for $2$ and $3$. There is no least upper bound. Similarly there is no greatest lower bound for $4$ and $5$. –  Jay Nov 13 '12 at 18:23

2 Answers 2

up vote 3 down vote accepted

The problem is that $2$ and $3$ have $2$ "smallest possible upperbounds" but they are not comparable so they don't have a join. Similarly, $4$ and $5$ don't have a meet.

Hope that helps,

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Remember for example that $4\wedge 5$ will be the unique greatest lower bound for 4 and 5. Now 2 and 3 are both lower bounds, so $4\wedge 5 $ above those.. but how could that be possible? There isn't such a node...

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