ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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The block sequence $1/1;\ 1/2, 2/2;\ 1/3, 2/3, 3/3;\ \dots$ has subsequential limit set exactly $[0,1]$

Example

Write $\iota(n) = n \cdot 1_{\mathbb{R}}$ for the canonical natural. Let $c : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$ be the function supplied by the recursion theorem (The recursion theorem) from the starting element $(1,1)$ and the function

$F(m, j) := (m,\, j+1) \ \text{ if } j < m, \qquad F(m, j) := (m+1,\, 1) \ \text{ if } j = m,$

write $c_k = (m_k, j_k)$ , and define the sequence of reals

$x_k \;:=\; \frac{\iota(j_k)}{\iota(m_k)} \qquad (k \in \mathbb{N}).$

Its first terms are

$\frac{1}{1};\quad \frac{1}{2}, \frac{2}{2};\quad \frac{1}{3}, \frac{2}{3}, \frac{3}{3};\quad \frac{1}{4}, \dots$

the blocks being the successive values of $m$ . Then the subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is the whole unit interval (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length),

$\operatorname{SL}(x) = [0,1],$

and consequently $\limsup_k x_k = 1$ and $\liminf_k x_k = 0$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ ).

The recursion replaces the block bookkeeping. Presenting the sequence by the partial sums $1 + 2 + \dots + (m-1)$ that mark where each block begins would require inverting that count at every index. Carrying the pair $(m, j)$ along instead makes each term's block and position immediately available, and the three facts the argument needs, that $1 \le j_k \le m_k$ , that $m_k \le k+1$ , and that every admissible pair occurs, are then three short inductions.

Facts & Assumptions

Given: The recursion $c_k = (m_k, j_k)$ described above, the sequence $x_k = \iota(j_k)/\iota(m_k)$ , and a real number $y$ with $0 \le y \le 1$ .

[L1]

Recursion theorem: for a set $A$ , an element $a \in A$ and $f : A \to A$ there is a unique $g : \mathbb{N} \to A$ with $g_0 = a$ and $g_{k+1} = f(g_k)$ (The recursion theorem, The natural numbers $\mathbb{N}$ (von Neumann)).

[L2]

Induction principle (The principle of mathematical induction).

[L3]

Well-ordering principle: every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L4]

Index maps: if $n_i < n_{i+1}$ for every $i$ then $n$ is strictly increasing, and then $n_i \ge i$ ; the composite $(x_{n_i})$ is a subsequence, and $L \in \operatorname{SL}(x)$ means some subsequence converges to $L$ (A strictly increasing index map satisfies $n_k \ge k$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences, Subsequential limit of a real sequence, and the subsequential limit set, Limits and Cauchy sequences of reals).

[L5]

Canonical naturals: $\iota(n) > 0$ and invertible for $n \ge 1$ , $\iota$ is strictly increasing, $\iota(1) = 1$ , and $\iota(n+n') = \iota(n) + \iota(n')$ (Canonical naturals are positive and strictly increasing).

[L6]

Order arithmetic: claim 4 of Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities and Inverses of positives are positive, and reciprocation reverses order state the strict forms, that multiplication by a positive element preserves $<$ , that inequalities may be translated and added, and that $0 < u < v$ gives $0 < 1/v < 1/u$ ; adjoining the case of equality, where the two sides coincide, gives the nonstrict forms used below. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives, and the order is total (Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

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

[L8]

Limits preserve non-strict inequalities (Limits preserve non-strict inequalities); a convergent sequence is bounded and a sequence diverging to $\pm\infty$ is unbounded (Every convergent sequence is bounded, Divergence to $+\infty$ and to $-\infty$ ).

[L9]

The interval $[0,1] = \{t \in \mathbb{R} : 0 \le t \le 1\}$ , with least element $0$ and greatest element $1$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L10]

[L11]

Absolute value: $|t| \le \eta$ if and only if $-\eta \le t \le \eta$ ; and the order on $\mathbb{N}$ is total with $k < k+1$ (Basic properties of the absolute value, Absolute value in an ordered field, Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ , Discreteness: $\sigma(n)$ is the immediate successor).

Verification

technique · constructive

1.1

The recursion theorem applied to $A = \mathbb{N} \times \mathbb{N}$ , the element $(1,1)$ and the function $F$ gives a unique $c$ with $c_0 = (1,1)$ and $c_{k+1} = F(c_k)$ , so $(m_k, j_k)$ and hence $x_k$ are well defined once $m_k \ge 1$ is known.

givenL1

1.2

For every $k$ one has $1 \le j_k \le m_k$ : this holds at $k = 0$ , where $(m_0, j_0) = (1,1)$ ; and if $1 \le j_k \le m_k$ then either $j_k < m_k$ , so that $c_{k+1} = (m_k, j_k + 1)$ with $1 \le j_k + 1 \le m_k$ , or $j_k = m_k$ , so that $c_{k+1} = (m_k + 1, 1)$ with $1 \le 1 \le m_k + 1$ . The claim follows by induction on $k$ ; in particular $m_k \ge 1$ always, so $\iota(m_k) > 0$ and $x_k$ is defined.

givenL2L5L11

1.3

For every $k$ one has $m_k \le k+1$ : at $k = 0$ , $m_0 = 1$ ; and $m_{k+1}$ is $m_k$ or $m_k + 1$ , so $m_{k+1} \le m_k + 1 \le k+2$ . This is again an induction on $k$ .

givenL2L11

1.4

For every natural $m \ge 1$ and every $j$ with $1 \le j \le m$ there is $k$ with $(m_k, j_k) = (m, j)$ . Induct on $m$ . For $m = 1$ the only admissible $j$ is $1$ , realised at $k = 0$ . Assume the claim for $m$ and apply it at $j = m$ to get $k$ with $(m_k, j_k) = (m, m)$ ; then $c_{k+1} = (m+1, 1)$ , and a second induction on $i$ shows $c_{k+1+i} = (m+1, 1+i)$ for every $i \le m$ : it holds at $i = 0$ , and if it holds at $i < m$ then $1 + i < m + 1$ , so $c_{k+2+i} = (m+1, 2+i)$ . Hence every $j$ with $1 \le j \le m+1$ is realised in block $m+1$ .

givenL1L2L11

1.5

Let $y$ be a real with $0 \le y \le 1$ .

givenL9

2.1

For every $k$ one has $0 < x_k \le 1$ : from $1 \le j_k \le m_k$ we get $0 < \iota(j_k) \le \iota(m_k)$ , and dividing by $\iota(m_k) > 0$ gives $0 < x_k \le 1$ .

step 1.2L5L6

2.2

For every natural $m \ge 1$ the set $\{\, j \in \mathbb{N} : 1 \le j \le m \text{ and } y\,\iota(m) \le \iota(j) \,\}$ is nonempty, since $j = m$ belongs to it because $y \le 1$ gives $y\,\iota(m) \le \iota(m)$ ; let $J(m)$ be its least element. Then $y \le \iota(J(m))/\iota(m)$ , and moreover $\iota(J(m))/\iota(m) \le y + 1/\iota(m)$ : if $J(m) = 1$ this reads $1/\iota(m) \le y + 1/\iota(m)$ , true because $y \ge 0$ ; and if $J(m) > 1$ then $j := J(m) - 1$ satisfies $1 \le j \le m$ and $j < J(m)$ , so minimality gives $\iota(j) < y\,\iota(m)$ , that is $\iota(J(m)) - 1 < y\,\iota(m)$ , and dividing by $\iota(m) > 0$ gives the bound. Hence $0 \le \iota(J(m))/\iota(m) - y \le 1/\iota(m)$ .

step 1.5L3L5L6L11

3.1

For every $L \in \operatorname{SL}(x)$ one has $0 \le L \le 1$ , since a subsequence of $(x_k)$ converging to $L$ satisfies $0 \le x_{n_i} \le 1$ at every index by step 2.1, and limits preserve non-strict inequalities. So $\operatorname{SL}(x) \subseteq [0,1]$ .

step 2.1L4L8L9

3.2

For every natural $m \ge 1$ the set $\{\, k \in \mathbb{N} : (m_k, j_k) = (m, J(m)) \,\}$ is nonempty by step 1.4, since $1 \le J(m) \le m$ ; let $K(m)$ be its least element. Then $x_{K(m)} = \iota(J(m))/\iota(m)$ , so $|x_{K(m)} - y| \le 1/\iota(m)$ by step 2.2, and $m = m_{K(m)} \le K(m) + 1$ by step 1.3.

step 1.4step 1.3step 2.2L3L11

4.1

Define $f : \mathbb{N} \to \mathbb{N}$ by $f(n) := K(n+2)$ ; then $f(n) \ge (n+2) - 1 = n+1 > n$ and $|x_{f(n)} - y| \le 1/\iota(n+2)$ . The recursion theorem applied to $\mathbb{N}$ , the element $K(1)$ and the function $f$ gives $n : \mathbb{N} \to \mathbb{N}$ with $n_0 = K(1)$ and $n_{i+1} = f(n_i)$ ; it is strictly increasing, so $n_i \ge i$ , and $|x_{n_{i+1}} - y| \le 1/\iota(n_i + 2) \le 1/\iota(i+2)$ .

step 3.2L1L4L5L6construct

5.1

The subsequence $(x_{n_i})$ converges to $y$ : given a real $\varepsilon > 0$ , take a natural $p \ge 1$ with $1/p < \varepsilon$ ; every $i \ge p$ satisfies $i \ge 1$ , so step 4.1 applied at $i-1$ gives $|x_{n_i} - y| \le 1/\iota(i+1)$ , and $i + 1 > p$ gives $1/\iota(i+1) < 1/p < \varepsilon$ . Hence $y \in \operatorname{SL}(x)$ , and since $y$ was an arbitrary element of $[0,1]$ , $[0,1] \subseteq \operatorname{SL}(x)$ .

step 4.1step 1.5L4L5L6L7L11

6.1

Therefore $\operatorname{SL}(x) = [0,1]$ . The sequence is bounded by step 2.1, so every subsequence of it is bounded and none diverges to $\pm\infty$ ; hence $\overline{\operatorname{SL}}(x) = \operatorname{SL}(x) = [0,1]$ , whose greatest element is $1$ and least element $0$ , and [L10] gives $\limsup_k x_k = 1$ and $\liminf_k x_k = 0$ .

step 5.1step 3.1step 2.1L8L9L10discharge-construct∎

Remarks

Every point of $[0,1]$ is approached, and the rate is the block width. In block $m$ the terms are $1/m, 2/m, \dots, m/m$ , spaced $1/m$ apart and covering $(0,1]$ , so any target in $[0,1]$ has a term of block $m$ within $1/m$ of it. Since blocks of every width occur, and occur arbitrarily late, this produces a subsequence converging to the target. That is the whole idea; steps 2.2 and 3.2 only make the choice of term canonical, by taking a least element rather than an arbitrary one, so that no choice principle is used.
The set is closed, as it must be. $[0,1]$ contains the limit of every convergent sequence of its own points, which is what If each $y_j$ is a subsequential limit of $(x_k)$ and $y_j \to y \in \mathbb{R}$ , then $y$ is a subsequential limit of $(x_k)$ predicts for any subsequential limit set. This example shows the prediction is not vacuous: the set here is an entire interval, in contrast with the two-point set of $(-1)^k$ has $\liminf = -1$ and $\limsup = 1$ , so it does not converge.
Neither endpoint is a value of the sequence in the case of $0$ . Every term is $> 0$ by step 2.1, so $0 \in \operatorname{SL}(x)$ is a genuine limit and not an attained value, while $1$ is attained, once in every block. Subsequential limits need not be values, and values need not be subsequential limits.
Why the sequence is not written by a closed formula. The classical presentation defines $x_k$ by first solving $m(m-1)/2 \le k < m(m+1)/2$ for $m$ , which needs a least-element argument at every index and a quadratic estimate to get $m \le k+1$ . The recursion carries the block and position forward instead, and the estimate $m_k \le k+1$ becomes the one-line induction of step 1.3.

Depends on

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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., §6.4 (standard reference, not scraped)

The block sequence 1/1; 1/2,2/2; 1/3,2/3,3/3; …1/1;\ 1/2, 2/2;\ 1/3, 2/3, 3/3;\ \dots1/1; 1/2,2/2; 1/3,2/3,3/3; … has subsequential limit set exactly [0,1][0,1][0,1]