Recursively defining the set of bit strings set having more zeros than ones Question: 
Recursively define the set of bit strings that have more zeros than ones.
I tried it this way:
$\Sigma\subset \{0,1\}^*$
Basis step: $0 \in \Sigma$
Recursive step: For any $x\in \Sigma$, $00x1\in L$
Is it a valid answer?

To expand on your recursive step, I believe you'll need:
$0x \in S$, $x0 \in S, 01x\in S, 10x\in S, 1x0\in S, 0x1\in S, x01\in S, x10\in S$, for any $x \in \Sigma$ 
And your $\Sigma$ should really denote the set you're trying to define, that is the set of bit strings having more zeros than ones.
