TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum

Statement

Let $(x_{k})$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let $S = {x_{k} : k \in N}$ be its range.

If $(x_{k})$ is nondecreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and $S$ is bounded above (Lower bound, bounded below, bounded set), then $(x_{k})$ converges (Limits and Cauchy sequences of reals) and $\lim_{k} x_{k} = \sup S .$
If $(x_{k})$ is nonincreasing and $S$ is bounded below, then $(x_{k})$ converges and $\lim_{k} x_{k} = \inf S .$

Both suprema and infima exist under the stated hypotheses: $S$ is nonempty, so the least-upper-bound property (Complete ordered field (least-upper-bound property)) supplies the first and Every nonempty set bounded below has an infimum the second, and each is unique (Suprema and infima are unique).

This is the first place in this track where the least-upper-bound property produces a limit. Everything on the sequences page held in any Archimedean ordered field; the theorem below does not, and the sequence of decimal truncations of $\sqrt{2}$ inside $Q$ is the standard witness of its failure there.

Statement references

Sequences of reals: bounded, eventually, frequently, tails, subsequences

Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences

Lower bound, bounded below, bounded set

Limits and Cauchy sequences of reals

Complete ordered field (least-upper-bound property)

Every nonempty set bounded below has an infimum

Suprema and infima are unique

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals with range $S = {x_{k} : k \in N}$ , which is nonempty since $x_{0} \in S$ .

[L1]

Least-upper-bound property and uniqueness: a nonempty subset of $R$ that is bounded above has a unique supremum, which is an upper bound of it (Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L2]

Greatest-lower-bound property and uniqueness: a nonempty subset of $R$ that is bounded below has a unique infimum, which is a lower bound of it (Every nonempty set bounded below has an infimum, Suprema and infima are unique).

[L3]

Epsilon characterisation of the supremum: if $u$ is an upper bound of a nonempty $S$ , then $u = \sup S$ exactly when for every $ε > 0$ there is $s \in S$ with $u - ε < s$ (Epsilon characterisation of the supremum).

[L4]

Epsilon characterisation of the infimum: if $ℓ$ is a lower bound of a nonempty $T$ , then $ℓ = \inf T$ exactly when for every $ε > 0$ there is $t \in T$ with $t < ℓ + ε$ (Epsilon characterisation of the infimum).

[L5]

Monotonicity: nondecreasing means $x_{j} \leq x_{k}$ whenever $j \leq k$ , and nonincreasing means $x_{j} \geq x_{k}$ whenever $j \leq k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L6]

Convergence, tested against a real $ε$ : $(x_{k})$ converges to $x$ when for every rational $ε > 0$ there is $K$ with $∣ x_{k} - x ∣ < ε$ for all $k \geq K$ , and producing such a $K$ for every real $ε > 0$ establishes this, since every positive rational is a positive real (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Absolute value: for $c > 0$ , $∣ t ∣ < c$ exactly when $- c < t < c$ (Basic properties of the absolute value).

[L8]

Bounded above and bounded below, for a subset of $R$ (Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1givenL1L8

Assume $(x_{k})$ is nondecreasing and $S$ is bounded above. Since $S$ is nonempty, $u : = \sup S$ exists, is unique, and is an upper bound of $S$ .

1.2givenL2L8

Assume $(y_{k})$ is nonincreasing with range $T$ bounded below. Since $T$ is nonempty, $ℓ : = \inf T$ exists, is unique, and is a lower bound of $T$ .

2.1step 1.1L1

Every term satisfies $x_{k} \leq u$ , because $x_{k} \in S$ and $u$ is an upper bound of $S$ .

2.2step 1.2L2

Every term satisfies $ℓ \leq y_{k}$ , because $y_{k} \in T$ and $ℓ$ is a lower bound of $T$ .

2.3step 1.1L3choose

Let $ε > 0$ be an arbitrary real. By [L3] there is $s \in S$ with $u - ε < s$ , and every element of $S$ is a term, so $s = x_{K}$ for some $K \in N$ .

2.4step 1.2L4choose

Let $ε > 0$ be an arbitrary real. By [L4] there is $t \in T$ with $t < ℓ + ε$ , and $t = y_{L}$ for some $L \in N$ .

3.1step 2.3L5

For every $k \geq K$ we have $x_{K} \leq x_{k}$ by monotonicity, hence $u - ε < x_{K} \leq x_{k}$ .

3.2step 2.4L5

For every $k \geq L$ we have $y_{k} \leq y_{L}$ by monotonicity, hence $y_{k} \leq y_{L} < ℓ + ε$ .

4.1step 2.1step 3.1L7algebra

For every $k \geq K$ : subtracting $u$ from $u - ε < x_{k} \leq u$ gives $- ε < x_{k} - u \leq 0 < ε$ , so $∣ x_{k} - u ∣ < ε$ .

4.2step 2.2step 3.2L7algebra

For every $k \geq L$ : subtracting $ℓ$ from $ℓ \leq y_{k} < ℓ + ε$ gives $- ε < 0 \leq y_{k} - ℓ < ε$ , so $∣ y_{k} - ℓ ∣ < ε$ .

5.1step 4.1L6

The real $ε > 0$ was arbitrary and $K$ was produced from it, so $(x_{k})$ converges to $u = \sup S$ , which is claim 1.

5.2step 4.2L6

The real $ε > 0$ was arbitrary and $L$ was produced from it, so $(y_{k})$ converges to $ℓ = \inf T$ , which is claim 2.

6.1step 5.1step 5.2∎

Both claims are established, so a nondecreasing sequence bounded above converges to the supremum of its range and a nonincreasing sequence bounded below converges to the infimum of its range.

Remarks

Only one half is proved twice. Claim 2 could instead be deduced from claim 1 by reflection, since $(- y_{k})$ is nondecreasing and bounded above and $\inf T = - \sup (- T)$ (Every nonempty set bounded below has an infimum). The direct argument is written out because it is no longer, and because it puts Epsilon characterisation of the infimum to work in the place it was proved for, rather than routing an infimum statement through a supremum statement and a sign change.
The limit is the supremum of the range, not merely some upper bound. That identification is what Epsilon characterisation of the supremum supplies and it is the useful part of the theorem: it is how a limit is computed from a monotone construction, as in the recursive sequences of the examples page, rather than merely shown to exist.
Boundedness on the other side is automatic and is not a hypothesis. A nondecreasing sequence is bounded below by $x_{0}$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences), so "nondecreasing and bounded above" already means "nondecreasing and bounded"; this is what makes A monotone sequence converges if and only if it is bounded an equivalence rather than a one-sided statement.
Without the hypothesis of boundedness the conclusion fails completely, and fails in a describable way: a nondecreasing sequence that is not bounded above diverges to $+ \infty$ (A nondecreasing sequence that is not bounded above diverges to $+ \infty$ ), so a nondecreasing sequence either converges to the supremum of its range or runs away, with no third possibility.

A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources