Consider the set $\;A = \left\{\frac5{x−3}:x>3\right\}$. How do I prove that $0$ is the infimum?

Question

Consider the set $\;A=\left\{\dfrac{5}{x-3}:x>3\right\}$.

I want to find the infimum $a$ of $A$ or prove that it doesn't exist. I have found that $0$ is a lower bound of this function by doing this :

$x > 3$

$x - 3 > 0$

$\dfrac1{x-3}>0$

$\dfrac5{x-3}>0$

Thus, I have rigorously found that $0$ is a lower bound of $A$, but I still haven't proven that it is the greatest lower bound. I tried to do it by contradiction ( assume that there exists a real number $b$ such that $\;0<b<\dfrac5{x-3}\;)$ . However I have no idea what to do after. Can someone please explain how to prove that $0$ is the infimum of $A$.

Answer 1

$\inf A =0$ means that for any $b > 0$, $b$ is not an upper bound. So we can do two things:

1. direct: Show that for any $b > 0$ we can always find an $x > 3$ so that $0 < \frac 5{x-3} < b$.

2. contradiction: Assume there is a $b >0$ that is a lower bound. That means that $b \le \frac 5{x-3}$ for all $x > 3$. Assume that and get a contradiction.

I recommend 1, but you chose 2) so lets do it.

Assume that for all $x> 3$ we have $0 < b \le \frac 5{x-3}$ thus

$b(x-3) \le 5$
$x-3 \le \frac 5b$
$x \le \frac 5b +3$....... for ALL $x > 3$. Thus, $\frac 5b +3$ is an upper bound for all $x > 3$. But the reals aren't bound above! Just take any $x$ so that $x > \frac 5b +3$.

That's your contradiction.

....

By the way. To do it by 1)

We need to find an $x>3$ so that $\frac 5{x-3} < b$. That would be the case if and only if $x-3 > \frac 5b$ or iff $x> \frac 5b + 3$. And as $\frac 5b >0$ if $x > \frac 5b +3$ then $x > 3$ and we can always find an $x > \frac 5b+3$.

SO if $x > \frac 5b +3$ then $0 < \frac 5{x-3} < b$ and $b$ can not be a lower bound.

Answer 2

If you assume that there is $b>0$ such that $0<b<\frac{5}{x-3}$ for all $x>3$, solve $\frac{5}{x-3}=\frac{b}{2}$ and find that $x=\frac{10}{b}+3>3$. Then plugging this $x$ makes a contradiction

Edit.

Assume that there is $b>0$ such that $0<b<\frac{5}{x-3}$ for all $x>3$. Now observe taht $\frac{10}{b}+3>3$. This implies that $$0<b<\frac{5}{\frac{10}{b}+3-3}=\frac{b}{2}$$ and it is a contradiction

Answer 3

You could also just take $\lim\limits_{x\to\infty}\frac{5}{x-3}=0$. Or if you do not know what a limes is, then for every $n\in\mathbb N$, $n>3$, we have $5n+3>3$ and so $\frac{5}{(5n+3)-3}=\frac 1 n\in A$, which goes to $0$.