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?
Denote by $\Sigma$ the set of all finite binary strings, including the empty string $\oslash$. In the follwing $x$, $y$, $\ldots$ are variables for strings, and juxtaposition denotes concatenation.

Strings containing more zeros than ones  are called admissible. The language $L$ consisting of all admissible strings can be recursively generated as follows:

(i)$\quad 0\in L$,

(ii)$\quad x\,y\in L\ \rightarrow\ x\,0\,y,\ x\,0\,1\,y,\ x\,1\,0\,y\in L$.

Note that in (ii) $x$ or $y$ may be $=\oslash$. – It is obvious that all strings so generated are admissible. Conversely: The admissible strings of length $\leq2$ are $0$ and $0\,0$, and are indeed produced by (i) and (ii). Assume now that all admissible strings of length $\leq n$ are produced by (i) and (ii). If an admissible string $z$ has length $n+1$ and does not consist solely of zeros it contains $0\,1$ or $1\,0$ as a substring. It then can  be produced using (ii) and an admissible string $x\,y$ of length $n-1$.
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.