Find infimum and closure of $A$

Question

Let $$ \mathbf{A}=\left\{\frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}: \mathbf{m}, \mathbf{n} \geq \mathbf{3}\right\} $$

Now find :

1. $inf(A)$

2. $int(A)$ , interior points

3. $cl(A)$ , closure of $A$

With euclidean metric on $\mathbb{R}$

We have $int(A) =\emptyset $ because $A$ is countable. also if $m$ be fixed then

$$ lim _{n \to \infty} \frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}= \frac{2}{7}$$

and if $n$ be fixed we have :

$$ lim _{m \to \infty} \frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}= \frac{3}{5}$$

Also for $m=n $ we have :

$$ lim _{m \to \infty} \frac{\mathbf{5 m}}{12\mathbf{m}}= \frac{5}{12}$$

Answer 1

$$\dfrac{3m + 2n}{5m + 7n} = k$$ $$(3m + 2n) = k(5m + 7n)$$ $$(7k -2)n + (5k-3)m = 0$$ $$-\dfrac{7k-2}{5k-3}= \dfrac{m}{n} > 0$$ so $\dfrac{7k-2}{5k-3} < 0$ and $\dfrac{2}{7} \leq k \leq \dfrac{3}{5}$. You can prove that $A$ is dense in $[2/7, 3/5]$, so $inf(A) = \dfrac{2}{7}$, $int(A) =\{\}$ and $cl(A) = [2/7, 3/5]$.