Complete market without risk-neutral measure

Let $$\mathcal{M}$$ be a one-period model with $$\Omega=\{\omega_1,\omega_2\}$$ and $$S_t^0=1$$ for $$t=0,1$$.

Find a $$D$$ such that $$S^d$$, $$d=1,...,D$$ yields a complete market without a risk-neutral measure. Why does this not contradict the second fundamental theorem of asset pricing?

So I guess this is due to the fact that it second fundamental theorem of asset pricing needs the market to be arbitrage-free. However, I was not able to construct such a model. Can someone help?

Consider a one period model with one risk-free asset that yields r (in your case, r=0 since $$S_t^{0}=1$$ for t=0,1), and one risky asset, so D=1. Now, this market is complete if and only if every contingent claim C is attainable (i.e, hedgeable). Let $$C(w_1)$$ and $$C(w_2)$$ the payoff of the Contingent claim in states $$w_1$$ and $$w_2$$, respectively. Let's call $$a:=S_1^1(w_1)$$ and $$b:=S_1^1(w_2)$$ To build a portfolio $$\xi_0$$, $$\xi_1$$ that replicates this contingent claim, we have to solve the system of equations:
$$(1+r)\xi_0 + \xi_1a = C(w_1)$$ $$(1+r)\xi_0 + \xi_1b = C(w_2)$$
This system has a unique solution when the determinant is not 0, i.e when $$(1+r)b - (1+r)a \ne 0$$, that is, when $$(1+r) \ne 0$$ and $$b \ne a$$.
Then, if $$b>a$$ and $$(1+r) \ne 0$$, the market is complete. However, if $$S_0^1 < a$$ (or $$S_0^1 > b$$), the market has an arbitrage opportunity, by buying $$S^1$$ (resp, selling $$S^1$$), borrowing from the risk-free asset and selling at t=1.