TITLE: Proving the sup and inf of $S$
QUESTION [1 upvotes]: Let there be a set of all numbers in the form $2^{-p}+3^{-q}+5^{-r}$ where $p,q,r$ each take on all positive integer (say this is set $S$). I know that the $\mbox{sup } S = \frac{31}{30}$ and $\mbox{inf } S = 0$, but how would I go about proving this?

REPLY [0 votes]: Well, $0 < 2^{-p} \le \frac 12$ and $0 < 3^{-q} \le \frac 13$ and $0 < 5^{-r} < \frac 15$ so
$0 <  2^{-p}+3^{-q}+5^{-r} \le \frac 12 + \frac 13 + \frac 15=\frac{31}{30}$
So $E = \{2^{-p}+3^{-q}+5^{-r} \}$ is bounded below by $0$ and above by $\frac{31}{30}$
As $\frac{31}{30} \in E$ then any $y < \frac{31}{30}$ is not an upper bound so $\sup E = \frac {31}{30}$.  (If you haven't proven it yet you should prove that if an upper bound, $s$, of $S$ is an element of $s$ then $\sup S = \max S = s$.)
$0 \not \in E$ so we can't make that argument.
But if $y > 0$ can we prove $y$ is not a lower bound?  In other words can we prove if $y > 0$ then there exist $p, q,r\in \mathbb N$ so that $2^{-p} +3^{-q} + 5^{-r} < y$?
And we can.
Let's see... we can safely assume $1 > y> 0$.  Let $N > \log_2 \frac 1y > 0$.  If $p > N$ then $2^p > \frac 1y$ and $y > 2^{-p} $.
Now let $y_1 = y- 2^{-p} > 0$.  Let $N_1 > \log_3 \frac 1{y_1}$ so if $q > N_1$ we have $y_1 > 3^{-q} $.
Let $y_2 = y_1 - 3^{-q} = y - 2^{-p} - 3^{-q} > 0$.  Let $N_2 > \log_5 \frac 1{y_2}$ so if $r > N_2$ we have $y_2 > 5^{-r} $
So $y_2 - 5^{-r} = y - 2^{-p} - 3^{-q}-5^{-r} > 0$ and $y >  2^{-p}+ 3^{-q}+5^{-r}$.
So $y$ is not a lower bound.
So $0 = \inf E$.
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