# Find the supremum, infimum, maximum and minimum

Find the supremum, infimum, maximum and minimum of this set:
$$E = \{\frac{2^p}{5^q}:{p \over q} \in (1,2)\text{ and } q > 0\}$$

My thoughts:

1. there is no supremum because we can choose always greater $p$.
2. therefore, there is no maximum
3. the infimum is $0$ when q converges to $\infty$.
4. no minimum, because between 0 and $E_n$ there's always a rational number according to the archimedes principle (or the density of the rationals). And of course, $0$ is not a term of $E$

Am I right? If not, please correct me.

• 1 is wrong. Think about it a bit more. (Also tidy up your notation!) Nov 24, 2013 at 15:53
• Why is it wrong? Nov 24, 2013 at 16:19
• Because $p$ has to be less than $2q$. Nov 24, 2013 at 16:23
• While it may be tempting, don't think about limits and convergence. You only need to use the definitions of supremum and infimum.
– user100690
Nov 24, 2013 at 16:25

Hint

For each $q$,

$$q<p<2q$$ because we want $\displaystyle \frac{p}{q}$ to be in the interval $(1,2)$. Thus,

$$p=2q-\epsilon$$

$\epsilon$ is a positive integer (Why?). Then

$$\frac{2^p}{5^q}=\frac{2^{2q-\epsilon}}{5^q}=\left(\frac{4}{5}\right)^q\left(\frac{1}{2}\right)^{\epsilon}$$.

You can conclude that this is less than $1$ (Why?). Hence your first conclusion may be faulty.

I think you should be able to make progress from here on.

• I get it now. so ${{4 \over 5}}$ must be the supremum of E Nov 24, 2013 at 16:41

The difference between supremum and maximum is that a maximum is in your set, while a supremum is the smallest upper bound. Since $2$ is an upper bound for $(1,2)$, every nonempty subset of $(1,2)$ has an upper bound and therefore (by properties of $\mathbb{R}$) has a supremum.