Applying Zorn's lemma to a set of sets of certain measures Thank you for reading. I am trying to work through a problem. While I have been given an indication that everything else I have done is in the right place, I cannot figure out the step to complete the proof. It is as follows, rephrased here as a problem in itself:
Let $S:= \{X \in \Sigma \ | \ \mu(X) \leq c \}$ for some countably additive finite measure $\mu$ over a $\sigma$-algebra $\Sigma$. Apply Zorn's lemma to demonstrate $S$ has a maximal element $s$ such that $\forall s' \in S$, $\mu(s) \geq \mu(s')$.
As an aside, I know separately that for any $\epsilon > 0$ and $A \in \Sigma$, $\exists B \subset A$ in the $\sigma$-algebra such that $0 < \mu(B) < \epsilon$, although I do not think this helps here.
Thank you for taking a look at this. Additionally note that proving $S$ is uncountably large would allow me to finish the proof a different way, though I am equally stumped as to whether this is even possible.
 A: We know $\Sigma$ is a sigma-algebra, and so it's closed under complementation and countable union. Let $C$ denote the collection of all chains in $S$ partially ordered by inclusion. We know that all of these chains are countably infinite at best, since they are elements of a sigma-algebra. Let $C_\alpha\in C,$ then $C_\alpha$ has the form
$$C_\alpha= X_{\alpha_1}\subset X_{\alpha_2}\subset \cdots \subset X_{\alpha_n}\subset\cdots.$$
Define $\mu(C_\alpha),$ to be the sequence of real numbers ordered in the same fashion: 
$$\mu(C_\alpha)= \mu X_{\alpha_1} \leq \mu X_{\alpha_2} \leq \cdots \leq \mu X_{\alpha_n} \leq \cdots \implies c_{\alpha_1}\leq c_{\alpha_2} \leq \cdots \leq c_{\alpha_n} \leq \cdots.$$
Thus, we have an increasing sequence of real numbers, bounded above by $c.$ So the monotone convergence theorem tells us that $\lim_{n\to\infty} c_{\alpha_n} = c_\alpha.$ 
Let $\{c_\alpha\}$ denote the set of all limits of sequences of this type. Then $\{c_\alpha\}\subset[0,c],$ and Bolzano Weierstrass tells us that this set $\{c_\alpha\}$ has a limit point, let us call it $c^\prime$. Let $\{c_{\alpha_n}^\prime\}$ be any sequence in $\{c_\alpha\}$ with this limit, then we take the chains associated to each $C_{\alpha_n},$ then you'll have to do some tedious arguing that you can build a new chain that is in $S,$ but it is doable. You may want to use a different definition of sigma-algebra which is equivalent "Any family of subsets of $X$ which is stable under differences and countable disjoint unions is a Sigma-algebra on $X.$"
