The supremum and infimum of the set $S_a = \{a/(a−x) \mid x \in (0, a)\}$

Question

Find the supremum and infimum of the set $S_a = \{a/(a−x) \mid x \in (0, a)\}$ for some $a>0$ and prove that you have found them.

For the supremum, I said that it did not exist, because the set goes to infinity as $x$ approaches $a$. For the infimum, I said that it was $1$. How do I prove these? For the supremum, can I just prove that the set is unbounded? Where can I start with proving the infimum?

Answer 1

Hint: Use characterization of infimum.

Show that $1$ is a lower bound and $1+\epsilon$ is not, for any choice of $\epsilon >0$

You're good for supremum. If a set is unbounded above, then it has no upper bound. Let alone, the lowest upper bound.

Answer 2

You're right about it being not bounded above, but you probably need to show that more rigorously.

Suppose we have a supremum $s=\frac{a}{a-b}$ for $b\in(0,a)$. Then let $c=\frac{a+b}{2}$. You should be able to show fairly easily $s'=\frac{a}{a-c}>s$, but $s'\in S_a$ which is a contradiction.

Also this idea of using $c=\frac{a+b}{2}$ should work as well for showing $1+\epsilon$ cannot be a lower bound for any $\epsilon$ (referencing the other answer).

As has been pointed out, what I said wasn't quite right. Instead, suppose we have an upper bound $M$ of the set. Then let $b=a-\frac{a}{M}$, so that

$$\frac{a}{M}=a-b$$ $$\frac{a}{a-b}=M$$

then you can do the same thing I said before.