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Every convergent sequence is bounded

Statement

Let $(x_{k})$ be a sequence of reals converging to $x \in R$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). Then $(x_{k})$ is bounded: there is $M \in R$ with $∣ x_{k} ∣ \leq M$ for every $k \in N$ .

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals converging to a real $x$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

Convergence: for every rational $ε > 0$ there is $K \in N$ with $∣ x_{k} - x ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals); and $1$ is a rational $> 0$ , since $Q$ is an ordered field (The rationals form a totally ordered field) and $0 < 1$ in every ordered field (The multiplicative identity is positive).

[L2]

Triangle inequality: $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ in $R$ (The triangle inequality, Complete ordered field (least-upper-bound property)).

[L3]

Absolute value: $∣ u ∣ \geq 0$ for every real $u$ (Basic properties of the absolute value).

[L4]

Finite maxima: for every $n \in N$ and all reals $a_{0}, \dots, a_{n}$ the set ${a_{0}, \dots, a_{n}}$ has a maximum (Every nonempty finite set of reals has a maximum and a minimum); a maximum lies in the set and dominates every element of it (Maximum and minimum of a set).

[L5]

Order arithmetic in $R$ : transitivity of $\leq$ and of $<$ ; both mixed forms, $u < v \leq w \Rightarrow u < w$ and $u \leq v < w \Rightarrow u < w$ , and the implication $u < v \Rightarrow u \leq v$ , all immediate from the reading of $u \leq v$ as " $u < v$ or $u = v$ " (Complete ordered field (least-upper-bound property), Ordered field); and translation invariance, $a < b \Rightarrow a + c < b + c$ (Order is preserved by adding a constant and by adding inequalities).

[L6]

The order on $N$ is total, so every index $k$ satisfies $k < K$ or $k \geq K$ ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1L1

Apply convergence with the rational $ε = 1$ : fix $K \in N$ such that $∣ x_{k} - x ∣ < 1$ for all $k \geq K$ .

2.1step 1.1L2L5

For every $k \geq K$ we have $∣ x_{k} ∣ = ∣ (x_{k} - x) + x ∣ \leq ∣ x_{k} - x ∣ + ∣ x ∣$ , while adding $∣ x ∣$ to both sides of $∣ x_{k} - x ∣ < 1$ gives $∣ x_{k} - x ∣ + ∣ x ∣ < 1 + ∣ x ∣ = ∣ x ∣ + 1$ ; composing the non-strict inequality with the strict one yields $∣ x_{k} ∣ < ∣ x ∣ + 1$ .

2.2step 1.1L4choose

Define $M : = \max {∣ x_{0} ∣, ∣ x_{1} ∣, \dots, ∣ x_{K - 1} ∣, ∣ x ∣ + 1}$ , the maximum of the finite list obtained by appending $∣ x ∣ + 1$ to the first $K$ values $∣ x_{0} ∣, \dots, ∣ x_{K - 1} ∣$ ; the list is nonempty because its last entry is always present (when $K = 0$ the list is the single entry $∣ x ∣ + 1$ ), so the maximum exists by [L4].

3.1step 2.2L4

For every $k < K$ the value $∣ x_{k} ∣$ is one of the entries of that list, hence $∣ x_{k} ∣ \leq M$ .

3.2step 2.1step 2.2L4L5

For every $k \geq K$ we have $∣ x_{k} ∣ < ∣ x ∣ + 1 \leq M$ , since $∣ x ∣ + 1$ is an entry of the list, hence $∣ x_{k} ∣ \leq M$ .

4.1step 3.1step 3.2L3L6∎

Every index satisfies $k < K$ or $k \geq K$ , and in both cases $∣ x_{k} ∣ \leq M$ ; therefore $(x_{k})$ is bounded.

Remarks

This is where Every nonempty finite set of reals has a maximum and a minimum does its main work on this page: the first $K$ terms are bounded because a nonempty finite set of reals has a maximum, and that fact is proved once, by induction, rather than re-derived here. It is not the only consumer, and no claim of uniqueness is made: A null sequence times a bounded sequence is null and Conventions for sequences: indexing, eventually, $\lim$ , and rational $ε$ cite the same lemma, each for a maximum of finitely many reals.
The converse is false. A bounded sequence need not converge (FALSE: every bounded sequence converges). What is true in that direction is the Bolzano-Weierstrass theorem, which requires a passage to a subsequence; it is not available at this point in the reading order, being the subject of the next page of this track, Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness, where it is proved.
Boundedness is exactly what makes A null sequence times a bounded sequence is null applicable to a convergent factor, and that is how boundedness enters the product rule of Algebra of limits: sums, scalar multiples, products and quotients.

Every convergent sequence is bounded

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources