Discussion
Let f be a mapping from X = {1, 2, 3, ......., 50} to itself such that for \(m, n \in X\), \(m \le n\) implies that \(f(m) \le f(n)\). Then which of the following is true ?
Answer: A
The true statement is there is \(m \in X\) such that \(f(m) = m\).
Consider \(g(m) = f(m) - m\). Since f maps into X, \(f(1) \ge 1\), so \(g(1) \ge 0\), and \(f(50) \le 50\), so \(g(50) \le 0\). Take the largest m with \(f(m) \ge m\). If \(f(m) \gt m\), then \(f(m) \ge m+1\), and because f is order-preserving \(f(m+1) \ge f(m) \ge m+1\), contradicting the choice of m. So \(f(m) = m\): every order-preserving map of a finite chain into itself has a fixed point.
The others fail. \(f(m) = m - 1\) for every m would give \(f(1) = 0\), outside X. \(f(m) = m + 1\) would give \(f(50) = 51\), outside X. And the identity map is order-preserving with \(f(m) = m\), so \(f(m) = \frac{1}{2}m\) is not forced for even m.
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