If $S = \varnothing$ then does an assumption $\forall A \in S: A \subset B \subset X \implies B \in S$ lead to a contradiction?

Question

Let $(X, \mathcal{T})$ be a topological space. My reading material defines a filter in $x$ as any set $S \subset P(X)$ s.t. 1.) $\varnothing \not\in S$, 2.) $A, B \in S \implies A \cap B \in S$, 3.) $\forall A\in S:A \subset B \subset X\implies B \in X$. The requirement of $S$ to be non-empty is not specified explicitly and I was wondering whether the situation $S = \varnothing$ would lead to a contradiction with the third assumption, i.e. if there are no elements in $S$ then does it mean, under the third assumption, that every subset of $X$ is part of $S$?

Answer 1

For any proposition or formula $P,$ the form $\forall x\in S\,(P)$ is an abbreviation for $\forall x\,(x\in S\implies P).$

And $(x\in S\implies P)$ is equivalent to $(\,(\neg x\in S)\lor P\,).$

So when $S=\emptyset$ we have $\forall x\in S\,(P)$ iff $\forall x\, (\,(\neg x\in \emptyset)\lor P \,).$ And $\forall x\,(\,(\neg x\in \emptyset)\lor P \,)$ is true regardless of what $P$ is.