Q:

Proving existence of maximal sum-free set containing $\{1,7,9\}$

I was asked to prove: 

Prove there exists a maximal sum-free set that contains the set $\{1,7,9\}$

We say a set of natural numbers is sum-free if no element in it is the sum of a finite number of other elements in the set, with no repetition of the elements in the summation.
and the hint was to use Zorn's lemma. 
So I defined a set $A=\{B\subseteq\mathbb{N}:\{1,7,9\}\subseteq B\text{ and } B \text{ is sum-free}\}$
And the order is inclusion, $\subseteq$.
Now I'd like to show that $A$ has a maximal element.
This can be obtained from Zorn's lemma if I can show that every chain in $A$ has an upper bound. This is where I'm stuck. 
Any idea on how to prove that every chain in $A$ has an upper bound? 
(I've got no intuition as to why such a maximal set should exist in the first place...)

A:

This is similar to the proof that the increasing union of functions is a function; or the increasing union of linearly independent sets is a linearly independent set.
If $C$ is a chain, and $B=\bigcup C$, then assume towards contradiction that $B$ is not sum-free, and derive that there was an element in $C$ which was already not sum-free.

It should be noted that in the case of $\Bbb N$, which can be enumerated, we can just proceed by induction and go over all the elements, then add at each step the elements which did not violate the sum-freeness of what previously was accumulated.
So there is no real need for Zorn's lemma here.

