Can a convergent sequence of real numbers have a subsequence converging to a different limit?
What does it mean for a sequence of functions to converge pointwise and to converge uniformly?
- Give an example of a sequence that converges pointwise but not uniformly.
Prove that every sequence admits a monotone subsequence.
Prove the monotone convergence theorem for sequences.
Prove the Bolzano-Weierstrass Theorem.
Solution
Functions \(f_n\colon E \to \RR\) converge pointwise to \(f\) if \(f_n(x) \to f(x)\) for every \(x \in E\), and converge uniformly to \(f\) if \(\sup_{x \in E}|f_n(x) - f(x)| \to 0\).
1. Every subsequence of a sequence converging to \(x\) converges to \(x\).
For \(\eps > 0\), all but finitely many terms lie within \(\eps\) of \(x\), and so do all but finitely many terms of any subsequence. Limits in \(\RR\) are unique.
2. \(f_n(x) = x^n\) on \([0,1]\) converges pointwise but not uniformly.
\(x^n \to 0\) for \(x \in [0,1)\) and \(1^n = 1\), so the pointwise limit is \(\chi_{\theset1}\), but \(\sup_{[0,1]}|x^n - \chi_{\theset1}(x)| = 1\) for every \(n\).
3. Every real sequence \((a_n)\) has a monotone subsequence.
Call \(n\) a peak if \(a_n \ge a_m\) for all \(m \ge n\). If there are infinitely many peaks \(n_1 < n_2 < \cdots\), then \(a_{n_1} \ge a_{n_2} \ge \cdots\). Otherwise let \(n_1\) exceed every peak. Since \(n_k\) is not a peak, there is \(n_{k+1} > n_k\) with \(a_{n_{k+1}} > a_{n_k}\), which gives an increasing subsequence.
4. An increasing sequence bounded above converges to its supremum, and an increasing sequence not bounded above tends to \(\infty\).
If \(s = \sup_n a_n < \infty\) and \(\eps > 0\), then \(s - \eps\) is not an upper bound, so \(a_N > s - \eps\) for some \(N\), and \(s - \eps < a_n \le s\) for \(n \ge N\). If \((a_n)\) is not bounded above, for every \(M\) some \(a_N > M\), and \(a_n \ge a_N > M\) for \(n \ge N\). Decreasing sequences are handled by passing to \((-a_n)\).
5. Every bounded real sequence has a convergent subsequence.
By step 3 it has a monotone subsequence, which is bounded and therefore converges by step 4.