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A subset of $R$ is compact iff it is sequentially compact

Statement

Let $K \subseteq R$ . Then $K$ is compact if and only if $K$ is sequentially compact (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

Neither implication is formal. Both are routed through the characterisation of compactness by closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded), and the forward implication additionally uses Bolzano-Weierstrass (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence). The backward implication uses the axiom of countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )): twice, once inside A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed when a point of the closure is turned into a sequence, and once directly in step 2.3, where an unbounded set supplies one point beyond each natural bound.

Facts & Assumptions

Given: A subset $K \subseteq R$ . Sequences are indexed by $N$ , which contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

$K$ is compact when every open cover has a finite subcover, and sequentially compact when every sequence with all terms in $K$ has a subsequence converging to a point of $K$ (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset, Subsequential limit of a real sequence, and the subsequential limit set, Limits and Cauchy sequences of reals).

[L2]

$K$ is compact exactly when $K$ is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded).

[L3]

Bolzano-Weierstrass: a sequence $(x_{k})$ of reals for which some $M$ satisfies $∣ x_{k} ∣ \leq M$ at every index has a subsequence converging to some real (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

[L4]

A point lies in $\overline{K}$ exactly when some sequence with all terms in $K$ converges to it, and $K$ is closed exactly when $\overline{K} = K$ , exactly when $K$ is sequentially closed (A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed, Interior, closure, boundary and exterior of a subset of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L5]

$K$ is bounded exactly when there are $ℓ, u$ with $ℓ \leq y \leq u$ for all $y \in K$ (Lower bound, bounded below, bounded set).

[L6]

Countable choice: for a family $(Y_{k})_{k \in N}$ of nonempty sets there is $f$ with domain $N$ and $f (k) \in Y_{k}$ for every $k$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L7]

A convergent sequence of reals is bounded (Every convergent sequence is bounded); every subsequence of a convergent sequence converges to the same limit (Subsequences inherit the limit); a sequence has at most one limit (A sequence has at most one limit); a strictly increasing $n : N \to N$ satisfies $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L8]

Archimedean property: for every real $z$ there is a natural $j \geq 1$ with $z < j$ ; canonical naturals satisfy $k \cdot 1_{R} \geq 0$ and are increasing in $k$ (Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing).

[L9]

Absolute value: $∣ z ∣ \geq z$ , $∣ z ∣ \geq - z$ , $∣ z ∣ \geq 0$ , and $∣ z ∣ = z$ for $z \geq 0$ while $∣ z ∣ = - z$ for $z < 0$ (Basic properties of the absolute value).

[L10]

Every nonempty finite set of reals has a maximum, which is one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1

For the forward implication assume $K$ is compact; then $K$ is closed and bounded by [L2], so [L5] supplies $ℓ, u$ with $ℓ \leq y \leq u$ for every $y \in K$ . Let $(x_{k})$ be any sequence with $x_{k} \in K$ for every $k \in N$ .

assume-hypL2L5

1.2

For the backward implication assume $K$ is sequentially compact.

assume-hypL1

2.1

The sequence of step 1.1 is bounded: put $M : = \max {∣ ℓ ∣, ∣ u ∣}$ by [L10]; for each $k$ , from $ℓ \leq x_{k} \leq u$ we get $x_{k} \leq u \leq ∣ u ∣ \leq M$ and $- x_{k} \leq - ℓ \leq ∣ ℓ ∣ \leq M$ , so $∣ x_{k} ∣ \leq M$ by [L9]. By [L3] there are a strictly increasing $n$ and a real $L$ with $x_{n_{j}} \to L$ ; every term $x_{n_{j}}$ lies in $K$ and $K$ is closed, so $L \in K$ by [L4]. Hence every sequence in $K$ has a subsequence converging in $K$ , that is, $K$ is sequentially compact.

step 1.1L1L3L4L9L10

2.2

A sequentially compact $K$ is closed: let $y \in \overline{K}$ ; by [L4] there is a sequence $(a_{k})$ with $a_{k} \in K$ for all $k$ and $a_{k} \to y$ ; by sequential compactness some subsequence $(a_{n_{j}})$ converges to a point $z \in K$ ; but that subsequence also converges to $y$ by [L7], and limits are unique by [L7], so $z = y$ and $y \in K$ . Hence $\overline{K} \subseteq K$ , so $\overline{K} = K$ and $K$ is closed by [L4].

step 1.2L1L4L7

2.3

A sequentially compact $K$ is bounded: suppose it is not. Then for every $k \in N$ the set $Y_{k} : = {y \in K : y > k or y < - k}$ is nonempty, since $Y_{k} = \emptyset$ would mean $- k \leq y \leq k$ for every $y \in K$ and make $K$ bounded by [L5]. Use [L6] to fix $f$ with $f (k) \in Y_{k}$ and put $x_{k} : = f (k)$ ; then $x_{k} \in K$ , and $∣ x_{k} ∣ > k$ for every $k$ , because $x_{k} > k \geq 0$ gives $∣ x_{k} ∣ = x_{k} > k$ while $x_{k} < - k \leq 0$ gives $∣ x_{k} ∣ = - x_{k} > k$ by [L9] and [L8]. By sequential compactness some subsequence $(x_{n_{j}})$ converges, hence is bounded by some real $M$ with $∣ x_{n_{j}} ∣ \leq M$ for all $j$ by [L7]; by [L8] fix a natural $j \geq 1$ with $M < j$ , and then $∣ x_{n_{j}} ∣ > n_{j} \geq j > M$ by [L7] and [L8], which contradicts $∣ x_{n_{j}} ∣ \leq M$ . So $K$ is bounded.

step 1.2L1L5L6L7L8L9

3.1

A sequentially compact $K$ is therefore closed by step 2.2 and bounded by step 2.3, hence compact by [L2].

step 2.2step 2.3L2

4.1

Step 2.1 is the forward implication and step 3.1 the backward one, so for subsets of $R$ compactness and sequential compactness coincide.

step 2.1step 3.1∎

Remarks

The equivalence is proved, not defined, and it is proved through the order. Both directions pass through A subset of $R$ is compact if and only if it is closed and bounded, whose backward half needs the completeness of $R$ , and the forward direction adds Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence, whose proof spends completeness again. Nothing here transfers to a setting where those are unavailable; see Which results on this page use the order of $R$ and therefore have no general-topological analogue.
Where the choices are spent, and whether they can be avoided. Step 2.3 selects one point of $K$ outside $[- k, k]$ for each $k$ , and A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed selects one point of $K$ in each shrinking neighbourhood. Both are countably many independent selections from subsets of $R$ , for which this library has no canonical rule, so The Axiom of Countable Choice ( ${A C}_{ω}$ ) is invoked rather than worked around. The forward implication, step 2.1, makes no such selection: the subsequence comes from Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence as a single object.
Sequential compactness is the form used in analysis; compactness is the form that is stated without sequences. The extraction of a convergent subsequence is what proofs about continuous functions on $[a, b]$ actually use, while the covering definition mentions no sequence and no limit. This theorem is what lets a reader move between them for subsets of $R$ , and it is proved only there.

Depends on

Used by

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sequentially compact space (Wikipedia) (standard reference, not scraped)
Bolzano-Weierstrass theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 and Ch. 3 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §7.4 (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

A subset of RR is compact iff it is sequentially compact