 We sketch a proof that \(\textrm{CH}\) implies the law of excluded middle. First we note that \(2^{\bN}\) is uncountable in the strong sense: Cantor’s diagonal argument shows that if \(S\) is a countable subset of \(2^{\bN}\), then there exists \(f \in 2^{\bN}\) such that \(f \notin S\). Assuming \(\textrm{CH}\), we now define  \[ X≡ \{S \in \textrm{dch}(\bN) : p \vee \neg p \}\]  and  \[ \textrm{C}≡\bN \cup X. \]  Then \(\bN ≲C≲\textrm{dch}(\bN)\). If \(C \cong \bN \), then \(p \vee \neg p \) is false, which is absurd. Hence \(\neg (C \cong \bN)\) and therefore \(C \cong \textrm{dch}(\bN)\). But \(\textrm{dch}(\bN)\), being equinumerous with \(2^{\bN}\) , is uncountable, so there exists \(S \in \textrm{dch}(\bN)\) that does not belong to the countable subset \(\bN\) of \(\textrm{dch}(\bN)\). Thus \(S \in C \), and therefore \(p \vee \neg p \) holds.