Discussion
Let \(f: \mathbb{N} \to \{1, 2, 3, \ldots, 100\}\) be an onto function. Consider the following statements.
P: If \(g(n) = \sup\{f(2n), f(2n+2), f(2n+4), \ldots\}\), \(n \geq 1\), then \(\lim_{n \to \infty} g(n) = \limsup_{n \to \infty} f(n)\).
Q: If \(h(n) = \inf\{f(n), f(n+1), f(n+2), \ldots\}\), \(n \geq 1\), then \(\lim_{n \to \infty} h(4n+1) = \liminf_{n \to \infty} f(n)\).
Then
Answer: B
P is NOT correct and Q is correct.
P fails because \(g(n)\) looks only at even arguments. \(\lim g(n)\) is the limit superior of the subsequence \(f(2), f(4), f(6), \ldots\), which can be smaller than \(\limsup f(n)\). For example, let \(f\) take the values 1 to 100 once each on \(n = 1, \ldots, 100\), then equal 1 at every even \(n\) and 100 at every odd \(n\) after that. It is onto, \(\limsup f(n) = 100\), but \(g(n) = 1\) for large \(n\).
Q holds. \(h(n)\) is the infimum of the tail from \(n\) onwards, so \(h\) is non-decreasing and bounded, and \(h(n) \to \liminf f(n)\) by definition. \(h(4n+1)\) is a subsequence of a convergent sequence, so it has the same limit.
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