Suppose $X$ is non-empty subset of non-negative $(x\geq0)$ elements in a field $\mathbb{F}$ which is bounded above. The sequence $(c_k)$ is defined as follows, Let $k$ be a natural number. $$b_k=\inf\{m \in \mathbb{N}: m \geq 2^k x, \forall x\in X\}, \,\,\, c_k=\frac{b_k}{2^k}$$ How to prove $\sup X = c$? I can prove $ \sup X \leq c$, $c$ is limit of the sequence.
My question is how to extend this idea for set that contains negative elements. If atleast one non-negative element is there, we can do the same thing (Is it?) . But how to do if all elements are negative? This method can not be used here I guess we get $\sup X=0$ if we use this method. Can we extend this idea to that or we should prove seperately?
Every bounded increasing sequence converges. That is the assumptions on the field.