Формулировка теоремы на языке сообщества математиков
t ⊢ a : N ↦ R ∧ b : N ↦ R ∧ c : N ↦ R ∧ ∃ N ∈ N ∀ n ∈ N ∧ n > N ( a n ≤ b n ≤ c n ) ∧ L ∈ R ∧ lim n → ∞ a n = L ∧ lim n → ∞ c n = L → lim n → ∞ b n = L {\displaystyle {\begin{aligned}t\vdash \quad \mathrm {a} :\mathbb {N} \mapsto \mathbb {R} \quad \land \quad \mathrm {b} :\mathbb {N} \mapsto \mathbb {R} \quad \land \quad \mathrm {c} :\mathbb {N} \mapsto \mathbb {R} \quad \land \quad \exists _{N\ \in \ \mathbb {N} }\ \forall _{n\ \in \ \mathbb {N} \ \land \ n\ >\ N}\ (a_{n}\leq b_{n}\leq c_{n})\\\ \land \quad L\in \mathbb {R} \quad \land \quad \lim _{n\to \infty }a_{n}=L\quad \land \quad \lim _{n\to \infty }c_{n}=L\quad \to \quad \lim _{n\to \infty }b_{n}=L\end{aligned}}}
Примечание
a : N ↦ R ⇔ a = { ⟨ n , a n ⟩ | ⟨ n , a n ⟩ ∈ N × R ∧ a n = a ( n ) } {\displaystyle ~\mathrm {a} :\mathbb {N} \mapsto \mathbb {R} \ \Leftrightarrow \ \mathrm {a} =\{\langle n,a_{n}\rangle |\ \ \langle n,a_{n}\rangle \in \mathbb {N} \times \mathbb {R} \ \land \ a_{n}=a(n)\}}
b : N ↦ R ⇔ b = { ⟨ n , b n ⟩ | ⟨ n , b n ⟩ ∈ N × R ∧ b n = b ( n ) } {\displaystyle ~\mathrm {b} :\mathbb {N} \mapsto \mathbb {R} \ \Leftrightarrow \ \mathrm {b} =\{\langle n,b_{n}\rangle |\ \ \langle n,b_{n}\rangle \in \mathbb {N} \times \mathbb {R} \ \land \ b_{n}=b(n)\}}
c : N ↦ R ⇔ c = { ⟨ n , c n ⟩ | ⟨ n , c n ⟩ ∈ N × R ∧ c n = c ( n ) } {\displaystyle ~\mathrm {c} :\mathbb {N} \mapsto \mathbb {R} \ \Leftrightarrow \ \mathrm {c} =\{\langle n,c_{n}\rangle |\ \ \langle n,c_{n}\rangle \in \mathbb {N} \times \mathbb {R} \ \land \ c_{n}=c(n)\}}
∃ N ∈ N ∀ n ∈ N ∧ n > N ( a n ≤ b n ≤ c n ) ⇔ ∃ N ( N ∈ N ∧ ∀ n ( n ∈ N ∧ n > N → a n ≤ b n ≤ c n ) ) {\displaystyle ~\exists _{N\ \in \ \mathbb {N} }\ \forall _{n\ \in \ \mathbb {N} \ \land \ n\ >\ N}\ (a_{n}\leq b_{n}\leq c_{n})\Leftrightarrow \exists N\ (N\in \mathbb {N} \ \land \ \forall n\ (n\in \mathbb {N} \ \land \ n>N\to a_{n}\leq b_{n}\leq c_{n}))}
Галактион 20:33, 25 червня 2009 (UTC)[відповісти]