Let \(X \subseteq \mathbb{N}\). Suppose \(X\) has the property that \(\forall n\in X \left[ n\subseteq X \right]\). Show that either \(X=\mathbb{N}\) or there exists \(n\in \mathbb{N}\) such that \(X = n\).
Proof. Let \(X\subseteq \mathbb{N}\), we consider \(\mathbb{N}\smallsetminus X\) which is also a subset of \(\mathbb{N}\). If \(\mathbb{N}\smallsetminus X\) is empty, then \(X = \mathbb{N}\), so go on vacation. Otherwise suppose \(\mathbb{N}\smallsetminus X\) non-empty, then by Lemma 4.11, \[\exists n\in \mathbb{N}, n\notin X \left[ (\mathbb{N}\smallsetminus X) \cap n = \emptyset \right].\]
Claim. \(X = n\).
Let \(x\in n\), and since \(n\in \mathbb{N}\), \(x \in n \subseteq \mathbb{N}\). If \(x \notin X\), then \(x \in \mathbb{N}\smallsetminus X\), contradicting fact that \(\mathbb{N}\smallsetminus X\) is empty, so \(x\in X\) and \(n \subseteq X\).
Let \(x\in X \subseteq \mathbb{N}\). Since \(x, n\in \mathbb{N}\), by Lemma 4.10 we have the trichotomy of either \(x = n\) or \(x \in n\) or \(n \in x\)
Case \(x=n\), then it implies \(n\in X\) which contradicts fact that \(n\notin X\).
Case \(n \in x\), by characterising property of \(X\) we have \(x\subseteq X\), so \(n\in X\), which again is a contradiction.
So \(x \in n\), then \(X \subseteq n\) so \(X = n\). \(\square\)