1. ## Infimum

If I use $x\leq\frac{k}{2^n}$ to show E is nonempty and bounded above, then E has a supremum. What would I use to show: If E is nonempty and bounded below, then E has an infimum. Would $x\geq\frac{k}{2^n}$ work? When I tried this the induction part doesn't make sense to me. For example by induction, then, there exists integers $k_n$ least in $A_n$ such that $k_o \leq \frac{k_1}{2} \leq \frac{k_2}{4} \leq ... \leq \frac{k_n}{2^n} \leq ...$ and to me that doesn't seem to make sense for something bounded below.

2. Will you please just state exactly what you are required to prove?
Is it, “If a set has a lower bound, then it has an infimum”?
If so, do you know that if a set has an upper bound them it has a supremum?

3. Yes... Prove that if E is nonempty and bounded below, then E has an infinum. Sorry about the confusion.

4. If you know that a set has an upper bound has a supremum, then the proof is simple.
Consider $F = \left\{ {x:\left( {\forall z \in E} \right)\left[ {x \leqslant z} \right]} \right\}$. That is $F$ is the set of all lower bounds for $E$.
Then $F$ is a nonempty set bounded above. Therefore $F$ has a supremum, $\gamma$.
Then, you show that $\gamma = \inf (E)$

5. How did you "use $k/2^n$" to prove that non-empty set, with upper bound, has a supremum? By using monotone convergence? You can do something similar by multiplying by -1 to make the increasing sequence $k/2^n$ into the decreasing sequence $-k/2^n$. Or do as Plato suggested and use the fact about supremum to prove infimum- again multiply by -1 to reverse the order.

6. Yes by using the monotone sequence property. This proof I'm working on atm does not specify that the set is bounded above or has a supremum.