I'm trying to teach myself category theory from Steve Awodey's [Category Theory](http://www.andrew.cmu.edu/course/80-413-713/notes/). Chapter 2 asserts:



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> It is not hard to see that a filter F is an ultrafilter just if for every element b ∈ B, either b ∈ F or ¬b ∈ F, and not both (exercise!).
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I've managed to prove the backwards implication, but the forwards implication is eluding me. I'm guessing the correct approach is to consider a filter F such that there exists a b ∈ B such that neither b ∈ F or ¬b ∈ F, and construct a superset filter F' which contains b, but I can't figure out how to construct F' and prove that it's a filter. Any hints much appreciated!



F is ultrafilter over a Boolean algebra implies that for every b, either b or not-b is in F?
Assume you have a proper filter $F$ that avoids both $b$ and $\neg b$. Then, you could consider the filter generated by $F\cup\{b\}$ - which is to say the smallest filter $F'$ containing $F$ and $b$. 


Since $F$ was a proper filter it follows that $0\not\in F$. 


If $0\in F'$, then this means that there is some $f\in F'$ such that $b\wedge f = 0$. Now, $\neg b=0\vee\neg b=(b\wedge f)\vee\neg b=(b\vee\neg b)\wedge(f\vee\neg b)=1\wedge(f\vee\neg b)=f\vee\neg b$. Thus $f≤\neg b$, which means that $\neg b\in F'$.


Since $\neg b\in F'$, either $\neg b\in F$ or $\neg b$ may be acquired by meets and upwards closures from $F\cup\{b\}$. Say $b\wedge f≤\neg b$ for some $f\in F$. Then $b\wedge f= b\wedge f\wedge\neg b = b\wedge\neg b\wedge f = 0\wedge f = 0$ for an $f\in F$ and by the above argument, we derive $\neg b\in F$. This is a contradiction, from which we can derive that $0\not\in F$.


Hence, $0\not\in F'$, and thus $F'$ is a proper ideal strictly containing $F$.


