§11 Subsequences

Why this matters

A sequence may wander without converging, yet still contain well-behaved pieces. Subsequences let us pull those pieces out. The payoff is the Bolzano–Weierstrass theorem below, which is the tool behind the existence of maxima and minima of continuous functions in §18.

11.1 Definition.

Suppose $(s_n)_{n\in\mathbb{N}}$ is a sequence. A subsequence of it is a sequence $(t_k)_{k\in\mathbb{N}}$ where for each $k$ there is a positive integer $n_k$ such that

$$n_1 < n_2 < \cdots < n_k < n_{k+1} < \cdots \tag{1}$$$$t_k = s_{n_k}. \tag{2}$$

Used in Exercise 11.6

So $(t_k)$ is a selection of some (possibly all) of the $s_n$’s, taken in order. Equivalently, (1) picks out an infinite subset $\{n_1, n_2, \dots\}$ of $\mathbb{N}$, and every infinite subset of $\mathbb{N}$ arises this way.

More formally, view $(s_n)$ as a function $s$ on $\mathbb{N}$. The function $\sigma(k) = n_k$ “selects” an infinite subset of $\mathbb{N}$ in order, and the subsequence is the composite $t = s \circ \sigma$:

$$t_k = t(k) = s(\sigma(k)) = s(n_k) = s_{n_k} \qquad\text{for } k \in \mathbb{N}. \tag{3}$$

So $t$ is a subsequence of $s$ if and only if $t = s \circ \sigma$ for some increasing $\sigma : \mathbb{N} \to \mathbb{N}$. We usually suppress $\sigma$ and $t$ and just say “a subsequence $(s_{n_k})$ of $(s_n)$.”

Common pitfallorder and no repeats

A subsequence keeps terms in their original order and uses each index at most once, because $n_1 < n_2 < \cdots$ is strict. Rearranging $(s_1, s_2, s_3, s_4, \dots)$ as $(s_2, s_1, s_4, s_3, \dots)$ does not give a subsequence, and neither does repeating a term as in $(s_1, s_1, s_2, s_2, \dots)$.

Example 1.

Let $s_n = n^2(-1)^n$, so $(s_n) = (-1, 4, -9, 16, -25, 36, \dots)$. Its positive terms form the subsequence $(4, 16, 36, 64, \dots)$, with $n_k = 2k$, $s_{n_k} = (2k)^2(-1)^{2k} = 4k^2$, and selection function $\sigma(k) = 2k$.

Example 2.

Let $a_n = \sin\frac{n\pi}3$. Then $(a_n) = \left(\frac{\sqrt3}2, \frac{\sqrt3}2, 0, -\frac{\sqrt3}2, -\frac{\sqrt3}2, 0, \frac{\sqrt3}2, \dots\right)$ and its subsequence of nonnegative terms is $\left(\frac{\sqrt3}2, \frac{\sqrt3}2, 0, 0, \frac{\sqrt3}2, \frac{\sqrt3}2, 0, 0, \dots\right)$, with $n_1, n_2, \dots = 1, 2, 3, 6, 7, 8, 9, 12, 13, \dots$ We don’t need a formula, but one is $n_k = k + 2\lfloor \frac k4 \rfloor$, where $\lfloor x \rfloor$ is the largest integer $\le x$.

The next theorem will seem obvious after some reflection, but it is good to have one careful proof that we can apply several times rather than repeat. (The first edition did repeat it.)

11.2 Theorem.

Let $(s_n)$ be a sequence.

1. If $t \in \mathbb{R}$, then there is a subsequence of $(s_n)$ converging to $t$ if and only if the set $\{n \in \mathbb{N} : |s_n - t| < \varepsilon\}$ is infinite for all $\varepsilon > 0$.
2. If $(s_n)$ is unbounded above, it has a subsequence with limit $+\infty$.
3. Similarly, if $(s_n)$ is unbounded below, a subsequence has limit $-\infty$.

In each case the subsequence can be taken to be monotonic.

Used in Theorem 11.7, Theorem 11.9, Exercise 11.7, Exercise 14.12

Proof

The forward implications are easy: for example, if $\lim_k s_{n_k} = t$ and $\varepsilon > 0$, then all but finitely many $n_k$ lie in $\{n : |s_n - t| < \varepsilon\}$. We prove the converses.

(i) If $\{n : s_n = t\}$ is infinite, there is a subsequence with $s_{n_k} = t$ for all $k$, which is a boring monotone sequence converging to $t$. So assume $\{n : s_n = t\}$ is finite. Then $\{n : 0 < |s_n - t| < \varepsilon\}$ is infinite for all $\varepsilon > 0$. These sets are unions $\{n : t - \varepsilon < s_n < t\} \cup \{n : t < s_n < t + \varepsilon\}$ that shrink as $\varepsilon \to 0$, so

$$\{n \in \mathbb{N} : t - \varepsilon < s_n < t\} \text{ is infinite for all } \varepsilon > 0, \tag{1}$$

or

$$\{n \in \mathbb{N} : t < s_n < t + \varepsilon\} \text{ is infinite for all } \varepsilon > 0; \tag{2}$$

otherwise both sets would be finite for small $\varepsilon$. Assume (1); case (2) is similar. We construct step by step a subsequence with $t - 1 < s_{n_1} < t$ and

$$\max\left\{s_{n_{k-1}}, t - \tfrac1k\right\} \le s_{n_k} < t \qquad\text{for } k \ge 2. \tag{3}$$

Such a subsequence is increasing, and $t - \frac1k \le s_{n_k} < t$ gives $\lim_k s_{n_k} = t$ (compare Exercise 8.5(a)).

Select $n_1$ with $t - 1 < s_{n_1} < t$, possible by (1). Having chosen $n_1 < \cdots < n_{k-1}$ satisfying (3), apply (1) with $\varepsilon = t - \max\{s_{n_{k-1}}, t - \frac1k\}$: infinitely many $n$ have $s_n$ in that range, so we can choose $n_k > n_{k-1}$ satisfying (3). This “definition by induction” produces the whole subsequence.

(ii) Let $n_1 = 1$. Given $n_1 < \cdots < n_{k-1}$, select $n_k > n_{k-1}$ with $s_{n_k} > \max\{s_{n_{k-1}}, k\}$. This is possible because $(s_n)$ is unbounded above, and only finitely many terms come before $n_{k-1}$. The result is monotone with limit $+\infty$. (iii) is similar.

∎

Intuitionlimits versus subsequential limits

Compare the two conditions on the interval $(t - \varepsilon, t + \varepsilon)$:

• $t$ is the limit when, for every $\varepsilon > 0$, the interval contains the terms $s_n$ for all but finitely many $n$;
• $t$ is a subsequential limit when, for every $\varepsilon > 0$, it contains $s_n$ for infinitely many $n$ (Theorem 11.2(i)).

So a subsequential limit is a place that the sequence keeps coming back to, though it may leave in between.

Example 3.

The rational numbers can be listed as a sequence $(r_n)$. One way: arrange $0, \pm\frac{m}{k}$ in an array and sweep through it diagonally, allowing repetitions. (Readers who know some set theory will recognize “$\mathbb{Q}$ is countable.”) This sequence has an amazing property: every real number $a$ is the limit of some subsequence of $(r_n)$. Indeed every interval $(a - \varepsilon, a + \varepsilon)$ contains infinitely many rationals (Exercise 4.11), so Theorem 11.2 applies.

Here is the diagonal sweep, step by step:

Cells swept12

BackNext

List so far: 0, 1/1, 1/2, 2/1, 3/1, 1/3, 1/4, 2/3, 3/2, 4/1, 5/1. Cells like 2/2 repeat an earlier value and are skipped; negatives are added by listing ±. Every fraction m/k sits on diagonal m + k, so it is reached after finitely many steps.

Example 4.

Suppose $s_n > 0$ for all $n$ and $\inf\{s_n\} = 0$. Then $(s_n)$ need not converge or even be bounded, but it has a subsequence converging monotonically to $0$. By Theorem 11.2 it suffices that $\{n : s_n < \varepsilon\}$ is infinite for each $\varepsilon > 0$. If it were finite for some $\varepsilon_0$, then either it is empty, so $\inf s_n \ge \varepsilon_0 > 0$, or $\inf s_n = \min\{s_n : s_n < \varepsilon_0\} > 0$. Either way we contradict $\inf s_n = 0$.

The next theorem is almost obvious.

11.3 Theorem.

If $(s_n)$ converges, then every subsequence converges to the same limit.

Used in Theorem 12.1, Theorem 13.5, Theorem 18.1

Proof

First, $n_k \ge k$ for all $k$: $n_1 \ge 1$, and $n_{k-1} \ge k - 1$ implies $n_k > n_{k-1} \ge k - 1$, so $n_k \ge k$. Let $s = \lim s_n$ and $\varepsilon > 0$. Choose $N$ with $|s_n - s| < \varepsilon$ for $n > N$. Then $k > N$ implies $n_k > N$, so $|s_{n_k} - s| < \varepsilon$.

∎

Intuitiona quick test for divergence

Read Theorem 11.3 backward: if two subsequences have different limits, the sequence cannot converge. This is often the fastest proof of divergence. For $\cos\frac{n\pi}{3}$, the subsequence $n_k = 6k$ is constantly $1$ and $n_k = 6k + 3$ is constantly $-1$. The converse direction is where people slip: one convergent subsequence says nothing about the whole sequence. $(-1)^n$ has a constant subsequence and still diverges.

Our immediate goal is the Bolzano–Weierstrass theorem: every bounded sequence has a convergent subsequence. First, monotone subsequences.

11.4 Theorem.

Every sequence $(s_n)$ has a monotonic subsequence.

Used in Theorem 11.5, Exercise 17.11

Proof

Call the $n$-th term dominant if it is greater than every term that follows it:

$$s_m < s_n \quad\text{for all } m > n. \tag{1}$$

Case 1: there are infinitely many dominant terms. Any subsequence consisting of dominant terms is decreasing by (1).

Case 2: there are only finitely many dominant terms. Select $n_1$ beyond all of them. Then

$$\text{given } N \ge n_1 \text{ there exists } m > N \text{ with } s_m \ge s_N, \tag{2}$$

since $s_N$ is not dominant. Apply (2) with $N = n_1$ to get $n_2 > n_1$ with $s_{n_2} \ge s_{n_1}$; then with $N = n_2$ to get $n_3$; and so on. By induction we obtain an increasing subsequence.

∎

This elegant proof comes from D. J. Newman’s book A Problem Seminar.

11.5 Bolzano–Weierstrass Theorem.

Every bounded sequence has a convergent subsequence.

Used in Theorem 13.5, §18, Theorem 18.1

Proof

A bounded sequence has a monotonic subsequence by Theorem 11.4, and that subsequence converges by Theorem 10.2.

∎

The Bolzano–Weierstrass theorem is very important and is used at critical points in Chapter 3. Our proof is a bit nonstandard. Convergence, Cauchy sequences and boundedness all make sense for sequences in the plane, but monotonicity does not. Bolzano–Weierstrass still holds there (Theorem 13.5), so in applications we emphasize 11.5 rather than 11.4.

Intuitiona second proof, by halving

Here is another standard proof of Bolzano–Weierstrass. Its pattern of repeated halving returns in the compactness arguments of §13. Suppose every $s_n$ lies in $[a, b]$. Cut the interval in half. At least one half contains $s_n$ for infinitely many $n$; keep that half. Cut it in half again and keep a half that still contains infinitely many terms, and so on. The kept intervals are nested and their lengths $\frac{b - a}{2^k}$ shrink to $0$, so by the nested interval property (§4) they close in on a single point $x$. Choose $n_1 < n_2 < \cdots$ with $s_{n_k}$ in the $k$th interval; this is possible because each interval contains infinitely many terms. Then $|s_{n_k} - x| \le \frac{b - a}{2^k}$, so $s_{n_k} \to x$. Unlike Theorem 11.4, this proof carries over to $\mathbb{R}^k$: there you cut a box into $2^k$ smaller boxes instead, exactly as in the proof of Proposition 13.13.

Subsequential limits

11.6 Definition.

Let $(s_n)$ be a sequence in $\mathbb{R}$. A subsequential limit is any real number, or symbol $+\infty$ or $-\infty$, that is the limit of some subsequence of $(s_n)$.

If $(s_n)$ has a limit $s$, then $\{s\}$ is the set of subsequential limits. The interesting case is when $(s_n)$ has no limit.

The sequence $s_n = \sin n$ is bounded but looks erratic, with no obvious pattern. Does it have a convergent subsequence?

Yes, even though you cannot write one down easilyNo, because the sequence itself does not converge

Try it: pick a subsequence

Choose a sequence and a selection rule $n_k$. The highlighted points are $s_{n_k}$. Try $\cos(n\pi/3)$ with $n_k = 6k$ and $n_k = 6k + 1$, or $\sin n$, where no simple rule finds a convergent subsequence even though Bolzano–Weierstrass guarantees one.

Terms shown60

sₙsubsequence s_(nₖ)

lim sup = 1 and lim inf = −1, so the sequence diverges (Theorem 10.7). The even terms converge to 1 and the odd terms to −1.

Choose your own terms

Now pick the $n_k$ yourself. Select any increasing list of indices and see whether the chosen terms settle toward one of the subsequential limits. The red and blue lines are $\limsup$ and $\liminf$, the largest and smallest subsequential limits (§12).

Even nOdd nEvery 3rdn ≡ 1 mod 6Clear

Choose the terms of your subsequence (n₁ < n₂ < …)

123456789101112131415161718192021222324252627282930313233343536

selected terms s_(nₖ)lim sup = 1lim inf = -1other subsequential limits

✗ The later selected terms are spread over a range of 1.5, so this selection does not settle down. It is not a convergent subsequence (within this window).

The six values repeat, so each value is a subsequential limit (pick every sixth term from the right starting point). The set is {−1, −1/2, 1/2, 1}.

Tab to a term’s checkbox and press Space to add or remove it, or click a point on the plot. Verdicts look only at the first 36 terms.

Example 5.

For $s_n = n^2(-1)^n$, the even terms diverge to $+\infty$ and the odd terms to $-\infty$. Every subsequence that has a limit diverges to $\pm\infty$, so $\{-\infty, +\infty\}$ is the set of subsequential limits.

Example 6.

$a_n = \sin\frac{n\pi}3$ takes each of the values $\frac{\sqrt3}2$, $0$, $-\frac{\sqrt3}2$ infinitely often. The only convergent subsequences are eventually constant, so $\{-\frac{\sqrt3}2, 0, \frac{\sqrt3}2\}$ is the set of subsequential limits. For instance, $n_k = 3k$ gives the constant $0$, $n_k = 6k + 1$ gives $\frac{\sqrt3}2$, and $n_k = 6k + 4$ gives $-\frac{\sqrt3}2$.

Example 7.

If $(r_n)$ lists all rationals, every real number is a subsequential limit (Example 3), and so are $\pm\infty$ (Exercise 11.7). So the set of subsequential limits is $\mathbb{R} \cup \{-\infty, +\infty\}$.

Example 8.

Let $b_n = n[1 + (-1)^n]$, so $b_n = 2n$ for even $n$ and $b_n = 0$ for odd $n$. The set of subsequential limits is $\{0, +\infty\}$.

Now the connection with $\limsup$ and $\liminf$.

11.7 Theorem.

Let $(s_n)$ be any sequence. There is a monotonic subsequence whose limit is $\limsup s_n$, and a monotonic subsequence whose limit is $\liminf s_n$.

Used in Theorem 11.8, Theorem 12.1, Theorem 14.9, Exercise 14.12

Proof

If $(s_n)$ is unbounded above, Theorem 11.2(ii) gives a monotone subsequence with limit $+\infty = \limsup s_n$. Similarly for unbounded below and $\liminf$.

Otherwise, consider the case that $(s_n)$ is bounded above (the other is similar), so $t = \limsup s_n$ is finite. Let $\varepsilon > 0$. There is $N_0$ with $\sup\{s_n : n > N\} < t + \varepsilon$ for $N \ge N_0$; in particular $s_n < t + \varepsilon$ for $n > N_0$. We claim

$$\{n \in \mathbb{N} : t - \varepsilon < s_n < t + \varepsilon\} \text{ is infinite.} \tag{1}$$

Otherwise there is $N_1 > N_0$ with $s_n \le t - \varepsilon$ for $n > N_1$. Then $\sup\{s_n : n > N\} \le t - \varepsilon$ for $N \ge N_1$, so $\limsup s_n < t$, a contradiction. Since (1) holds for every $\varepsilon > 0$, Theorem 11.2(i) gives a monotone subsequence converging to $t$.

∎

11.8 Theorem.

Let $(s_n)$ be any sequence in $\mathbb{R}$ and let $S$ be its set of subsequential limits.

1. $S$ is nonempty.
2. $\sup S = \limsup s_n$ and $\inf S = \liminf s_n$.
3. $\lim s_n$ exists if and only if $S$ has exactly one element, namely $\lim s_n$.

Used in §5, §10, §12, §14

Proof

(i) follows from Theorem 11.7.

(ii) Let $t$ be the limit of a subsequence $(s_{n_k})$. By Theorem 10.7, $t = \liminf_k s_{n_k} = \limsup_k s_{n_k}$. Since $n_k \ge k$, $\{s_{n_k} : k > N\} \subseteq \{s_n : n > N\}$ for each $N$, so

$$\liminf_n s_n \le \liminf_k s_{n_k} = t = \limsup_k s_{n_k} \le \limsup_n s_n.$$

This holds for all $t \in S$, so $\liminf s_n \le \inf S \le \sup S \le \limsup s_n$. By Theorem 11.7, $\liminf s_n$ and $\limsup s_n$ belong to $S$, so (ii) holds.

(iii) is a reformulation of Theorem 10.7.

∎

So $\limsup s_n$ is exactly the largest subsequential limit and $\liminf s_n$ the smallest. That makes them easy to compute.

Example 9.

If $s_n = n^2(-1)^n$, then $S = \{-\infty, +\infty\}$, so $\limsup s_n = +\infty$ and $\liminf s_n = -\infty$.

Example 10.

If $a_n = \sin\frac{n\pi}3$, then $S = \{-\frac{\sqrt3}2, 0, \frac{\sqrt3}2\}$, so $\limsup a_n = \frac{\sqrt3}2$ and $\liminf a_n = -\frac{\sqrt3}2$.

Example 11.

If $(r_n)$ lists all rationals, then $\limsup r_n = +\infty$ and $\liminf r_n = -\infty$.

Example 12.

If $b_n = n[1 + (-1)^n]$, then $\limsup b_n = +\infty$ and $\liminf b_n = 0$.

The set of subsequential limits always contains the limits of its own convergent sequences. Such sets are called closed and are studied further in §13.

11.9 Theorem.

Let $S$ be the set of subsequential limits of $(s_n)$. If $(t_n)$ is a sequence in $S \cap \mathbb{R}$ and $t = \lim t_n$, then $t \in S$.

Used in Exercise 11.9

Proof

Suppose $t$ is finite and let $\varepsilon > 0$. Some $t_n$ lies in $(t - \varepsilon, t + \varepsilon)$. Let $\delta = \min\{t + \varepsilon - t_n, t_n - t + \varepsilon\}$, so $(t_n - \delta, t_n + \delta) \subseteq (t - \varepsilon, t + \varepsilon)$. Since $t_n$ is a subsequential limit, $\{m : s_m \in (t_n - \delta, t_n + \delta)\}$ is infinite, hence so is $\{m : s_m \in (t - \varepsilon, t + \varepsilon)\}$. By Theorem 11.2(i), $t \in S$.

If $t = +\infty$, then $(s_n)$ is unbounded above, so $+\infty \in S$ by Theorem 11.2(ii). Similarly for $-\infty$.

∎

Reading alongside Abbott

Abbott’s Section 2.5 proves Bolzano–Weierstrass by the halving argument above, using the nested interval property. Ross’s proof through dominant terms (Theorem 11.4) gives something stronger on the line, a monotone subsequence, but it does not carry over to $\mathbb{R}^k$, which is why Ross gives a separate argument in Theorem 13.5. Subsequential limits as a set (Theorems 11.7–11.9) get much more attention here than in Abbott, and they lead into §12.

Check your understanding

Take the quiz (3 questions)

Which of these is a subsequence of $(s_n)$?

$(s_2, s_4, s_6, \dots)$$(s_1, s_1, s_2, s_2, s_3, s_3, \dots)$$(s_2, s_1, s_4, s_3, \dots)$

What is the set of subsequential limits of $s_n = (-1)^n + \frac1n$?

$\{-1, 1\}$$\{-1, 0, 1\}$$[-1, 1]$

Which sequence must have a convergent subsequence?

$s_n = \sin n$$s_n = n$$s_n = (-1)^n n$

Exercises

Exercise 11.1

Let $a_n = 3 + 2(-1)^n$ for $n \in \mathbb{N}$.

1. List the first eight terms of $(a_n)$.
2. Give a subsequence that is constant (takes a single value). Specify the selection function $\sigma$.

Solution

(a) $1, 5, 1, 5, 1, 5, 1, 5$. (b) $\sigma(k) = 2k$ gives the constant subsequence $5, 5, 5, \dots$; $\sigma(k) = 2k - 1$ gives $1, 1, 1, \dots$

Tags: Warm-upSubsequences

Exercise 11.2

Consider $a_n = (-1)^n$, $b_n = \frac1n$, $c_n = n^2$, $d_n = \frac{6n+4}{7n-3}$.

1. For each sequence, give an example of a monotone subsequence.
2. For each sequence, give its set of subsequential limits.
3. For each sequence, give its lim sup and lim inf.
4. Which of the sequences converges? diverges to $+\infty$? diverges to $-\infty$?
5. Which of the sequences is bounded?

Solution

(b) $\{-1, 1\}$, $\{0\}$, $\{+\infty\}$, $\{\frac67\}$. (c) $(a_n)$: $1, -1$; $(b_n)$: $0, 0$; $(c_n)$: $+\infty, +\infty$; $(d_n)$: $\frac67, \frac67$. (d) $(b_n)$ and $(d_n)$ converge; $(c_n)$ diverges to $+\infty$; none diverges to $-\infty$. (e) $(a_n)$, $(b_n)$, $(d_n)$.

Tags: Warm-upSubsequences

Used in Exercise 11.3, Exercise 11.4

Exercise 11.3

Repeat Exercise 11.2 for $s_n = \cos\frac{n\pi}3$, $t_n = \frac{3}{4n+1}$, $u_n = \left(-\frac12\right)^n$, $v_n = (-1)^n + \frac1n$.

Solution

Subsequential limits: $\{-1, -\frac12, \frac12, 1\}$, $\{0\}$, $\{0\}$, $\{-1, 1\}$. lim sup / lim inf: $1/-1$, $0/0$, $0/0$, $1/-1$. $(t_n)$ and $(u_n)$ converge to $0$; all four are bounded.

Tags: Warm-upSubsequences

Exercise 11.4

Repeat Exercise 11.2 for $w_n = (-2)^n$, $x_n = 5^{(-1)^n}$, $y_n = 1 + (-1)^n$, $z_n = n\cos\frac{n\pi}4$.

Solution

Subsequential limits: $\{-\infty, +\infty\}$, $\{\frac15, 5\}$, $\{0, 2\}$, $\{-\infty, 0, +\infty\}$. lim sup / lim inf: $+\infty/-\infty$, $5/\frac15$, $2/0$, $+\infty/-\infty$. None converges or diverges to $\pm\infty$; $(x_n)$ and $(y_n)$ are bounded.

Tags: Warm-upSubsequences

Exercise 11.5

Let $(q_n)$ be an enumeration of all the rationals in the interval $(0, 1]$.

1. Give the set of subsequential limits for $(q_n)$.
2. Give the values of $\limsup q_n$ and $\liminf q_n$.

Solution

(a) $[0, 1]$. (b) $1$ and $0$.

Tags: StandardSubsequences

Exercise 11.6

Show every subsequence of a subsequence of a given sequence is itself a subsequence of the given sequence.

Hint

Define subsequences as in (3) of Definition 11.1: $s \circ \sigma \circ \tau$, and a composition of increasing maps $\mathbb{N} \to \mathbb{N}$ is increasing.

Solution

A subsequence of $(s_n)$ is a composite $s \circ \sigma$ with $\sigma : \mathbb{N} \to \mathbb{N}$ strictly increasing. A subsequence of that is $(s \circ \sigma) \circ \tau = s \circ (\sigma \circ \tau)$ with $\tau$ strictly increasing. Now $\sigma \circ \tau$ is strictly increasing: $k < k'$ gives $\tau(k) < \tau(k')$, and then $\sigma(\tau(k)) < \sigma(\tau(k'))$. So $s \circ (\sigma \circ \tau)$ is a subsequence of $(s_n)$.

Tags: Warm-upSubsequencesFrom the definitions

Exercise 11.7

Let $(r_n)$ be an enumeration of the set $\mathbb{Q}$ of all rational numbers. Show there exists a subsequence $(r_{n_k})$ such that $\lim_{k\to\infty} r_{n_k} = +\infty$.

Hint

$(r_n)$ is unbounded above. Apply Theorem 11.2(ii).

Solution

Every real number $M$ is exceeded by some rational number, and $(r_n)$ lists every rational, so $r_n > M$ for some $n$. Hence $(r_n)$ is unbounded above, and Theorem 11.2(ii) gives a subsequence $(r_{n_k})$ with $\lim r_{n_k} = +\infty$.

Tags: StandardSubsequences

Exercise 11.8

Use Definition 10.6 and Exercise 5.4 to prove $\liminf s_n = -\limsup(-s_n)$ for every sequence $(s_n)$. (This exercise is referred to in several places.)

Solution

For each $N$, apply Exercise 5.4 to the set $S_N = \{s_n : n > N\}$: $\inf S_N = -\sup(-S_N)$, and $-S_N = \{-s_n : n > N\}$. So $\inf\{s_n : n > N\} = -\sup\{-s_n : n > N\}.$ Let $N \to \infty$. The left side tends to $\liminf s_n$ (Definition 10.6). The right side is the negative of a sequence tending to $\limsup(-s_n)$, so it tends to $-\limsup(-s_n)$ (Theorem 9.2 with $k = -1$, or Exercise 9.10(b) in the infinite cases). Hence $\liminf s_n = -\limsup(-s_n)$.

Tags: StandardSubsequencesSup & inf

Used in §12, Exercise 12.5, Exercise 12.6

Exercise 11.9

1. Show the closed interval $[a, b]$ is a closed set.
2. Is there a sequence $(s_n)$ such that $(0, 1)$ is its set of subsequential limits?

Solution

(a) If $t_n \in [a, b]$ and $t_n \to t$, then $t \in [a, b]$ by Exercise 8.9(c). (b) No: by Theorem 11.9 the set of subsequential limits is closed, but $\frac1n \in (0,1)$ converges to $0 \notin (0, 1)$.

Tags: StandardOpen, closed, compactSubsequences

Exercise 11.10

Form an infinite array whose every row is $1, \frac12, \frac13, \frac14, \frac15, \dots$, and let $(s_n)$ list its entries by zig-zagging along the diagonals (the book’s Figure 11.2).

1. Find the set $S$ of subsequential limits of $(s_n)$.
2. Determine $\limsup s_n$ and $\liminf s_n$.

Solution

(a) Each $\frac1m$ occurs infinitely often, so $S = \{\frac1m : m \in \mathbb{N}\} \cup \{0\}$. (b) $\limsup s_n = 1$ and $\liminf s_n = 0$.

Tags: StandardSubsequences

Exercise 11.11

Let $S$ be a bounded set. Prove there is an increasing sequence $(s_n)$ of points in $S$ with $\lim s_n = \sup S$. Compare Exercise 10.7. (If $\sup S \in S$, it suffices to let $s_n = \sup S$ for all $n$.)

Solution

Let $s = \sup S$. If $s \in S$, take $s_n = s$ for all $n$. Otherwise, choose $s_1 \in S$ with $s_1 > s - 1$. Given $s_n \in S$, we have $s_n < s$, so $\max\{s_n, s - \frac1{n+1}\} < s$ and there is $s_{n+1} \in S$ larger than this maximum. Then $(s_n)$ is increasing, and $s - \frac1n < s_n \le s$ for all $n$, so $s_n \to s$.

Tags: StandardSup & infSubsequences

Used in Exercise 10.7, Exercise 13.13