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In the bounded real-valued functions on $N$ with the supremum metric, the closed unit ball is closed and bounded and is not compact: the indicator functions of the singletons are pairwise at distance $1$

Statement refuted

Refuted claim: in every metric space a closed and bounded subset is compact (FALSE: a closed and bounded subset of a metric space is compact).

The witness is the space $B (N)$ of bounded functions $N \to R$ with the supremum metric $d_{\infty} (f, g) = \sup {∣ f (s) - g (s) ∣ : s \in N}$ (The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set), together with the closed unit ball

$K : = {f \in B (N) : d_{\infty} (f, 0) \leq 1}$

about the zero function $0$ . The set $K$ is closed in $B (N)$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and it is not compact (Open cover, subcover, compact metric space, and compact subset of a metric space): the indicator functions $e_{n}$ of the singletons ${n}$ lie in $K$ and satisfy $d_{\infty} (e_{m}, e_{n}) = 1$ whenever $m \neq n$ , so no finite $1 / 2$ -net for $K$ can exist.

This is the same failure as in FALSE: a closed and bounded subset of a metric space is compact, in the space where it matters analytically: closed bounded sets of a function space are routinely not compact.

Facts & Assumptions

Given: The set $B (N)$ of bounded functions $N \to R$ with the supremum metric $d_{\infty}$ , the zero function $0$ , the closed unit ball $K$ , and for $n \in N$ the function $e_{n}$ with $e_{n} (n) = 1$ and $e_{n} (s) = 0$ for $s \neq n$ .

[L1]

$d_{\infty} (f, g) = \sup {∣ f (s) - g (s) ∣ : s \in N}$ is a metric on $B (N)$ , and the supremum is a member-free upper bound that is least among upper bounds (The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set, Lower bound, bounded below, bounded set, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

The closed ball $\overset{ˉ}{B} (x, r) = {y : d (x, y) \leq r}$ is closed, and a subset contained in a ball is bounded (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

A compact metric space is totally bounded; a subset is compact exactly when the metric subspace is; and a finite $ε$ -net is a finite subset whose $ε$ -balls cover the space (A compact metric space is complete and totally bounded, and neither implication uses any choice principle, Finite $ε$ -net and totally bounded metric space, Open cover, subcover, compact metric space, and compact subset of a metric space, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

$N$ is not equinumerous with any natural number, a subset of $N$ bounded above is finite, and an injection puts its domain in bijection with its image (The pigeonhole principle on $N$ , Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable, Injection, surjection, bijection, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The natural numbers $N$ (von Neumann)).

Counterexample

technique · direct

1.1

Each $e_{n}$ is a bounded function $N \to R$ , its range being contained in ${0, 1}$ , so $e_{n} \in B (N)$ ; and $d_{\infty} (e_{n}, 0) = \sup {∣ e_{n} (s) ∣ : s \in N} = 1$ , so $e_{n} \in K$ .

2.1

For $m \neq n$ the set ${∣ e_{m} (s) - e_{n} (s) ∣ : s \in N}$ is ${0, 1}$ , since the value is $1$ at $s = m$ and at $s = n$ and $0$ elsewhere; hence $d_{\infty} (e_{m}, e_{n}) = 1$ .

L1step 1.1

2.2

$K$ is the closed ball $\overset{ˉ}{B} (0, 1)$ , hence closed in $B (N)$ , and it is contained in $B (0, 2)$ , hence bounded.

L2step 1.1

2.3

Suppose $K$ had a finite $1 / 2$ -net $F$ ; then $F$ is empty or listable as ${y_{0}, \dots, y_{p}}$ , and it is not empty because $K$ contains $e_{0}$ . Define $φ : N \to σ (p)$ by letting $φ (n)$ be the least $i \leq p$ with $d_{\infty} (e_{n}, y_{i}) < 1 / 2$ , which exists because the $1 / 2$ -balls about the $y_{i}$ cover $K$ and $e_{n} \in K$ .

L3L4step 1.1

3.1

$φ$ is injective: if $φ (m) = φ (n) = i$ with $m \neq n$ , then $d_{\infty} (e_{m}, e_{n}) \leq d_{\infty} (e_{m}, y_{i}) + d_{\infty} (y_{i}, e_{n}) < 1 / 2 + 1 / 2 = 1$ , contradicting step 2.1.

L1step 2.1step 2.3

4.1

So $N$ is in bijection with $φ [N]$ , a subset of $N$ bounded above by $p$ and therefore finite, making $N$ equinumerous with a natural number, which is false.

L5step 3.1

5.1

Hence $K$ has no finite $1 / 2$ -net, so $K$ is not totally bounded and therefore not compact, while being closed and bounded by step 2.2; the claim of FALSE: a closed and bounded subset of a metric space is compact is refuted.

L3step 2.2step 4.1∎

Remarks

What goes wrong is room, not size. The ball $K$ has diameter $2$ , so it is small in the metric sense; but it contains infinitely many points that are pairwise at distance $1$ , so no finite family of small balls reaches all of them. That is exactly the failure of total boundedness, and it is one of the two conditions that, by For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, are together equivalent to compactness once the Axiom of Countable Choice and the Axiom of Dependent Choice are assumed.

What the argument does and does not settle. It shows that $K$ fails total boundedness, and that alone rules out compactness by [L3]. Nothing above is claimed about whether $K$ is complete; the point of the witness is that closedness and boundedness, the two conditions that suffice in $R^{n}$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), do not suffice here.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

73 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

Uniform norm (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)

In the bounded real-valued functions on NN with the supremum metric, the closed unit ball is closed and bounded and is not compact: the indicator functions of the singletons are pairwise at distance 11