# Order relation, Hasse diagram and its properties

Let $n \in \mathbb N$ and the function $\text{ld}(x)$ be mapping each number with its last digit (in decimal notation). Then consider the functions $f$ and $g$ defined as follows: $$f:n\in \mathbb N \to \text{r}(\text{ld}(n), 7)\;\;\;g:n\in \mathbb N \to \text{r}(\text{ld}(n), 13)$$ knowing that $\text{r}(a,b)$ is the function that returns the remainder from the division of $a$ by $b$.

Let $\rho$ be a order relation on the set $S:=\{n\in\mathbb N \mid n < 10\}$, then $$a \;\rho\; b \Leftrightarrow a = b \text{ or } (f(a)<f(b) \text{ and } g(a)<g(b))$$

I need to

• draw the Hasse diagram relative to $(S, \rho)$;
• tell if $(S, \rho)$ is a lattice;
• find $\{4,8\}$ lower bounds and $\{3,7\}$ upper bounds.

* My attempt to soultion *

I have drawn the following diagram:

is it possible that an element has no relation to any other in the set? In that case is it correct the way I have built the diagram?

Provided that $7$ has no relations, this isn't a lattice because we can't find a $\sup$ or an $\inf$.

I think the only lower bound for $\{4,8\}$ is $0$, but there are no upper bounds for $\{3,7\}$ as $7$ isn't relatable to any element in $S$.

Is my reasoning correct? Have I done anything wrong?

• Why doesn't $7\;\rho\; 1$? – Xodarap Jun 19 '13 at 15:58
• @Xodarap because $7 \neq 1$, $f(7) = 0 < f(1) = 1$ but $g(7)=7 \not{<} g(1) = 1$, so $7 \not{\rho} 1$. – haunted85 Jun 19 '13 at 16:02

$\sup$ or $\inf$ of what? It is clear what you mean, but anyway...