- Suppose that for each positive integer \(m\), we have a set \(S_m\) of real numbers, a real number \(\alpha_m\), and a real function \(f_m\). Formulate the negation of this statement: “There exists a positive integer \(m\) such that for every \(x \in S_m\), \(x \ge \alpha_m\) and \(f_m(x) = 0\).”
- The following statement is not valid: “For any positive integer \(m\), if \(T\) is a set of positive integers such that (1) \(m \in T\) and (2) \(n \in T\) implies \(n+1 \in T\), then \(T = \{\text{positive integers } n : n \ge m\}\).” Explain and correct the flaw.
Solution
For part (a), the statement is \[ \exists m\in\mathbb Z_{>0}\;\forall x\in S_m, \bigl(x\ge \alpha_m\ \text{and}\ f_m(x)=0\bigr). \] Negating successively gives \[ \forall m\in\mathbb Z_{>0}\;\exists x\in S_m \quad\text{such that}\quad x<\alpha_m\ \text{or}\ f_m(x)\ne0. \]
For part (b), the flaw is that conditions (1) and (2) force \(T\) to contain every integer \(n\ge m\), but they do not forbid \(T\) from also containing positive integers smaller than \(m\). For example, with \(m=3\), the set of all positive integers satisfies (1) and (2), but is not \(\{n:n\ge3\}\).
The correct conclusion from (1) and (2) is \[ \{n\in\mathbb Z_{>0}:n\ge m\}\subseteq T. \] Indeed, induction on \(r\ge0\) gives \(m+r\in T\). To obtain equality, add the hypothesis \[ T\subseteq\{n\in\mathbb Z_{>0}:n\ge m\}. \] Then the two inclusions give \[ T=\{n\in\mathbb Z_{>0}:n\ge m\}. \]