On Boolean algebras and Atoms

Question

I know plenety of questions have been made around this topics, but i've got an specific one. If $\mathfrak A$ Is an infinite boolean algebra, and $C$ is a finite subset of atoms in $\mathfrak A$, then $\sum _{a \in c} a \neq 1$

I was tryng this by induction on the size of $C$. The base case would be having two atoms $a$ and $b$. I supposed $a + b$ could not be $1$, because that would mean $\mathfrak A$ is finite. Then i suppose a subset $C$ of atoms having size $n$ fulfilling the requirement , and at the induction step i used a similar argument as in the base of the induction, saying that $\mathfrak A$ wold be finite if the $n$ atoms plus a new one gave me $1$ as result.

I'm not convinced at all, but i kind find any mistakes. Moreover if this proof is totally wrong can you give me a hint or idea on how to proceed.

Answer 1

The idea of the proof is good, but should be probably called 'proof by contradiction' instead of 'induction':

If we have $a_1+\dots+a_n=1$ for atoms $a_i$, then every element $x$ of the Boolean algebra is a sum of certain $a_i$'s (specifically, of those that are smaller than $x$).

Answer 2

Here's another approach.

Let $\mathbf B$ be an atomic Boolean Algebra (not all infinite Boolean Algebras are atomic) and $A$ be its set of atoms.
Prove that, for $a \in A$, $$\bigvee (A \setminus \{a\}) = a'.$$ It follows that if $A_0$ is a proper subset of $A$, then $\bigvee A_0 \leq a'<1$, for each $a \in A \setminus A_0$.