Show the properties of this relation! Is my solution correct?

Question

enter preformatted text hereLet $X = \mathbb{N}^+$ and define $\preccurlyeq$ relation on $X$ set:

$$x \preccurlyeq y \Leftrightarrow \exists k \in \mathbb{N} : y = kx$$

a.) Show that $\preccurlyeq$ is an (partial) order.

b.) Let $A=\{6, 10\}$. Give the smallest and minimum element/elements, if those exists.

c.) Show the infimum of $A$.

a.) $X=\mathbb{N}^+$ $$ x \preccurlyeq\ y \Leftrightarrow x\mid y $$ A relation is an partial order if it's reflexive, antisymmetric, and transitive.

• reflexive: $$x \mid x \quad x \leq x: \quad \exists k \in \mathbb{N} \quad x=k*x \quad k=1$$

• antisymmetric: $$x \leq y: \quad \exists k_1 \in \mathbb{N}: \quad y=k_1*x$$ $$y \leq x: \quad \exists k_2 \in \mathbb{N}: \quad x=k_2*y$$ $$y=k_1k_2*y, \quad where \quad k_1k_2=1 \quad and \quad k_1=k_2=1 \Rightarrow x=y$$

• transitive: $$ x \leq y: \quad \exists k_1 \in \mathbb{N}: \quad y=k_1*x$$ $$ y \leq z: \quad \exists k_2 \in \mathbb{N}: \quad z=k_2*y$$ $$z=|k_1k_2|*x, \quad where \quad |k_1k_2| \in \mathbb{N} \quad so \quad x \leq z$$

b.) $A=\{6, 10\}$

minimum element:

$$6 \leq 6 \quad \checkmark; \quad 6 \leq 10 \quad this \quad is \quad a \quad contradiction, \quad because \quad \nexists k \in \mathbb{N}: \quad 6*k = 10$$ $$10 \leq 6 \quad this \quad is \quad a \quad contradiction, \quad because \quad \nexists k \in \mathbb{N}: \quad 10*k = 6; \quad 10 \leq 10 \quad \checkmark$$

$\Rightarrow \nexists \quad minimum \quad element$

smallest element:

$$\nexists x: x \leq 6 \Leftarrow 10 \nleq 6, \quad because \quad \nexists k \in \mathbb{N}: 6=k*10 $$ $$\nexists x: x \leq 10 \Leftarrow 6 \nleq 10, \quad because \quad \nexists k \in \mathbb{N}: 10=k*6 $$

c.) infimum of $A$

$infA=gcd\left(6,10\right)$

$ 2 \leq 6;10 \quad 1 \leq 6;10$

Lower bounds: $1, 2$

$divisors \quad of \quad 6: 1, 2, 3 \qquad because \quad 1;2;3*k=6$ $divisors \quad of \quad 10: 1, 2, 5 \quad because \quad 1;2;5*k=10$

$\Rightarrow the \quad common \quad divisors \quad of \quad 6 \quad and \quad 10: 1, 2$

So only $1$ and $2$ could be the lower bounds of $A$ because $\nexists k_1 \in \mathbb{N}: 3*k_1=10$ and $\nexists k_2 \in \mathbb{N}: 5*k_2=6$

According to the definition of infimum:

$$1 \leq 2 \Rightarrow infA=2$$

Is my solution correct?

Answer 1

For a)

1. There's a typo in proving antisymmetry. It should be $x=k_1\cdot y$ instead of $x=k_2\cdot x$.
2. Maybe a word about how you get to $k_1=k_2=1$ from $k_1k_2=1$ would be nice
3. In proving transitivity, there's no real need for absolute values.

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For b) I don't really know what "smallest" means, and I don't know what you mean by "unit" (do you mean "element") but your proof there is no minimum element of $A$ is correct.

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For c)

You only proved that $2$ is a lower bound for $A$. You didn't prove that $1$ and $2$ are the only lower bounds for $A$, so you didn't yet prove that $2$ is the infimum (i.e., the largest lower bound).