Existence of non-satisfiable set of n propositions with all proper subsets satisfiable, for all n in N

Question

I'm trying to prove that for all $n \in N$, there is a set of $n$ propositions $\Sigma_n$ such that all of its proper subsets are satisfiable but $\Sigma_n$ itself is not satisfiable. All propositions are in propositional logic.

I tried two different approaches but neither seemed to work. First I tried to do induction on $n$, assuming there is a $\Sigma_{k}$ with the required properties, and let $\Sigma_{k} = \Sigma_{ksub\alpha} \cup \ ${$ \alpha $} for some arbitrary proposition $\alpha \in \Sigma_{k}$, and try to find two new propositions $\beta$ and $\gamma$ to add to $\Sigma_{k-1}$ to make it work, but I can't seem to continue from there.

I also noticed that the inductive hypothesis is equivalent to the statement that all proper subsets with $k-1$ elements, $\Sigma_{ksub\alpha}$, are satisfiable, and $\Sigma_{ksub\alpha} \models (\lnot \alpha)$. Since $\Sigma_{ksub\alpha} \models (\lnot \alpha)$ means either $\Sigma_{ksub\alpha} \ \cup \ ${$ (\lnot \alpha)$} or $\Sigma_{ksub\alpha}$ is not satisfiable, and that $\Sigma_{ksub\alpha}$ is satisfiable, it's same as suggesting $\Sigma_{ksub\alpha} \ \cup \ ${$ (\lnot \alpha)$} is satisfiable. Now the goal would be to show $\Sigma_{k+1sub\beta} \ \cup \ ${$ (\lnot \beta)$} is satisfiable (and same for $\gamma$). After changing the goal to prove I don't know how to proceed either.

Answer 1

I think you're overthinking this. Here's a hint: we can get an example in which all but one of the sentences are (distinct) propositional atoms, e.g. for $n=5$ there is an example of the form $$\{p_1, p_2, p_3, p_4, \varphi\}$$ where only $\varphi$ is more complicated than just a single propositional atom. Do you see how to do this?

Just take $\varphi=\neg(p_1\wedge p_2\wedge p_3\wedge p_4)$.

And this generalizes easily:

For any $n$, we can use the set $$\{p_1,...,p_{n-1}, \neg(p_1\wedge...\wedge p_{n-1})\}.$$