<s>[INST] Find infimum and closure of $A$ 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 :

*

*$inf(A)$


*$int(A)$ , interior points


*$cl(A)$ , 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 :
$$ lim _{m \to \infty} \frac{\mathbf{5 m}}{12\mathbf{m}}= \frac{5}{12}$$
 [/INST] $$\dfrac{3m + 2n}{5m + 7n} = k$$
$$(3m + 2n) = k(5m + 7n)$$
$$(7k -2)n + (5k-3)m = 0$$
$$-\dfrac{7k-2}{5k-3}= \dfrac{m}{n} > 0$$
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 $cl(A) = [2/7, 3/5]$.
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