TITLE: Free variables in first order logic
QUESTION [5 upvotes]: Let $\sigma = \{\subseteq\}$ be a signature with the binary relation symbol $\subseteq$. For a set $S$, we define the $\sigma-$structure $\mathcal{S} = (\mathcal{P(S), \subseteq^\mathcal{S}})$  with $\mathcal{P(S)}$ and $\subseteq^\mathcal{S}$ being the powerset of $S$ and the ordinary subset relation respectively. Give two formulas $\varphi_1, \varphi_2 \in \text{FO}[\sigma]$ such that:
$ \varphi_1(\mathcal{S}) = \emptyset.$
$\varphi_2(\mathcal{S}) = \{ x \in \mathcal{P(S)} \mid \text{ } \mid x \mid = 1 \}$.

What I understand is that our universe is a set all subsets of a set $S$. For $\varphi_1$ I have:
$$\varphi_1(x) = \forall y (x \subseteq y)$$
Since the empty set is a subset of every set including itself.
For $\varphi_2$:
$$\varphi_2(x) = \forall y (y \subseteq x \rightarrow x \subseteq y \lor \varphi_1(y))$$
Here I'm basically saying that $y$ and $x$ have to be the same set (with one element).
Can you please check my work?

REPLY [2 votes]: $\varphi_1$ is fine, but $\varphi_2$ actually defines $\{x\in\wp(\mathcal{S}):|x|\color{red}{\le}1\}$, since $\varnothing$ satisfies it. That’s very easily fixed:
$$\varphi_2(x)=\neg\varphi_1(x)\land\forall y\big(y\subseteq x\to x\subseteq y\lor\varphi_1(y)\big)$$