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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:

enter image description here

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?

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

Yes. A minor nitpick:

... this isn't a lattice because we can't find a sup or an inf.

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

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