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The sequence $(- 1)^{k} (1 + 1 / k)$ is bounded with subsequential limit set exactly ${- 1, 1}$

Example

For $k \geq 1$ let

$x_{k} = (- 1)^{k} (1 + \frac{1}{k}) .$

Then $(x_{k})$ is bounded, with $1 < ∣ x_{k} ∣ \leq 2$ at every index, it does not converge, and its subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is exactly

$SL (x) = {- 1, 1} .$

The example separates two things that a first reading of Bolzano-Weierstrass can run together. A bounded sequence must have a subsequential limit; it may have several; and having several is exactly what stops it converging. Here there are two, and neither is a value of the sequence, since $∣ x_{k} ∣ > 1$ always.

Indexing and the sign. Written on $N$ the sequence is $u_{j} : = t_{j} (1 + 1 / (j + 1))$ for $j \in N$ , where $t_{j} : = - s_{j}$ and $(s_{k})$ is the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ . Since $s_{σ (j)} = - s_{j}$ , the sequence $(t_{j})$ is $j \mapsto s_{j + 1}$ , so $u_{j} = x_{j + 1}$ and $(u_{j})$ is the family above under the substitution $k = j + 1$ . The verification uses $(u_{j})$ .

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ and the index maps $e, o$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; the sequence $t_{j} : = - s_{j}$ ; the sequence $p_{j} : = 1 + 1 / (j + 1)$ , where $j + 1$ denotes the canonical natural $(j + 1) \cdot 1_{R}$ ; and $u_{j} : = t_{j} p_{j}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{i}} = 1$ and $s_{o_{i}} = - 1$ for every $i$ , and $e$ and $o$ are strictly increasing (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , and $m \mapsto m \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L3]

Reciprocals: $a > 0$ gives $1 / a > 0$ , and $0 < a \leq b$ gives $0 < 1 / b \leq 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L5]

Absolute value: $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ , $∣ t ∣ \geq 0$ , $∣ t ∣ = t$ for $t \geq 0$ , and $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L6]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); subsequences inherit the limit (Subsequences inherit the limit); the absolute value is compatible with limits (The absolute value is compatible with limits); limits are unique (A sequence has at most one limit).

[L7]

Convergence and boundedness of a sequence of reals; it suffices to test a real $ε > 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L8]

Subsequential limits: $L \in SL (u)$ exactly when some subsequence of $(u_{j})$ converges to $L$ (Subsequential limit of a real sequence, and the subsequential limit set).

[L9]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1givenL1L2L3L5

For every $j$ : $j + 1 \geq 1$ gives $0 < 1 / (j + 1) \leq 1$ , so $1 < p_{j} \leq 2$ ; and $∣ t_{j} ∣ = ∣ - s_{j} ∣ = ∣ s_{j} ∣ = 1$ .

1.2givenL2L3L4L5L7

The sequence $(p_{j})$ converges to $1$ : given a real $ε > 0$ , [L4] supplies a natural $n \geq 1$ with $1 / n < ε$ , and for $j \geq n$ we have $0 < 1 / (j + 1) \leq 1 / n < ε$ , so $∣ p_{j} - 1 ∣ = 1 / (j + 1) < ε$ .

2.1step 1.1L5L7

$(u_{j})$ is bounded, with $1 < ∣ u_{j} ∣ = ∣ t_{j} ∣ p_{j} = p_{j} \leq 2$ at every index.

2.2step 1.2L1L6L8

Along the even index map: $t_{e_{i}} = - s_{e_{i}} = - 1$ , so $u_{e_{i}} = - p_{e_{i}}$ ; since $e$ is strictly increasing, $(p_{e_{i}})_{i}$ is a subsequence of $(p_{j})$ and so converges to $1$ , whence $u_{e_{i}} \to - 1$ .

2.3step 1.2L1L6L8

Along the odd index map: $t_{o_{i}} = - s_{o_{i}} = 1$ , so $u_{o_{i}} = p_{o_{i}} \to 1$ by the same argument.

3.1step 2.2step 2.3L8

Hence $- 1 \in SL (u)$ and $1 \in SL (u)$ .

3.2step 1.2step 2.1L6L8

Conversely, let $L \in SL (u)$ and fix a strictly increasing $n$ with $u_{n_{i}} \to L$ . Then $∣ u_{n_{i}} ∣ \to ∣ L ∣$ ; but $∣ u_{n_{i}} ∣ = p_{n_{i}}$ by step 2.1, and $(p_{n_{i}})_{i}$ is a subsequence of $(p_{j})$ , so it converges to $1$ .

4.1step 3.2L5L6L9

By uniqueness of limits $∣ L ∣ = 1$ , so $L = 1$ or $L = - 1$ .

5.1step 2.1step 3.1step 4.1L6L8∎

Combining, $SL (u) = {- 1, 1}$ ; and $(u_{j})$ does not converge, since a convergent sequence has exactly one subsequential limit. Bounded by step 2.1, the sequence $(u_{j})$ , that is $(x_{k})$ , therefore has the asserted properties.

Remarks

Where Bolzano-Weierstrass sits. Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence guarantees that a bounded sequence has at least one subsequential limit; this example computes the whole set and finds two. Nothing above uses the theorem, since the two subsequential limits are exhibited directly, and that is the honest order of business: an existence theorem is not needed once a witness is in hand.
The perturbation is what makes the example more than the alternating sequence. With $p_{j}$ replaced by the constant $1$ the sequence is $\pm 1$ alternating and each subsequential limit is attained infinitely often. Here $∣ x_{k} ∣ > 1$ at every index, so neither subsequential limit is ever a value of the sequence. Subsequential limits are limits, not values.
The converse direction is where the partition earns its keep. Step 4.1 rules out every other candidate by taking absolute values, which collapses the sign and leaves the single null perturbation to handle. The alternative, splitting an arbitrary subsequence according to how many of its indices are even, needs the disjointness half of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ and is longer.
This is the standard example of a bounded divergent sequence with a computable $lim sup$ and $lim inf$ , namely $1$ and $- 1$ . Those notions are developed on the next page of the track, and this item is written so that it can be cited there without change.

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Sources

Subsequential limit (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.6 (standard reference, not scraped)
MIT 18.100A, Lecture 9: Limsup, Liminf, and the Bolzano-Weierstrass Theorem (standard reference, not scraped)
J. Lebl, Basic Analysis I, Limit superior, limit inferior, and Bolzano-Weierstrass (standard reference, not scraped)

The sequence (−1)k(1+1/k)(−1)k(1+1/k) is bounded with subsequential limit set exactly {−1,1}{−1,1}