False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: in every normed space a closed bounded set is compact

Statement

False claim: in every normed space $(W,N)$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) a subset that is closed in the induced metric (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) is compact (Open cover, subcover, compact metric space, and compact subset of a metric space).

What is true is the same statement for $\mathbb{R}^{n}$ with the Euclidean norm and $n$ a natural number, which is Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ 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 clause 2. The published FALSE: a closed and bounded subset of a metric space is compact already refutes the corresponding claim for arbitrary metric spaces; the point of the present item is that adding a linear structure and a norm does not repair it, which a reader who has just met For $n \ge 1$ all norms on $\mathbb{R}^n$ are equivalent may well expect it to.

The witness is the space $V$ of finitely supported real sequences with the norm $N_\infty$ , both as in FALSE: all norms on a real vector space are equivalent, and the closed unit ball

$K \;:=\; \{\, v \in V : N_\infty(v) \le 1 \,\} .$

$K$ is closed in $(V, d_{N_\infty})$ and bounded, and it is not compact: the vectors $e_k$ all lie in it and satisfy $N_\infty(e_j - e_k) = 1$ for $j \ne k$ .

Facts & Assumptions

Given: The vector space $V$ of finitely supported sequences and the norm $N_\infty$ on it, with induced metric $d(v,w) := N_\infty(v-w)$ ; the set $K$ above; and the vectors $e_k \in V$ with $e_k(k) = 1$ and $e_k(j) = 0$ for $j \ne k$ .

[A1]

The refuted claim, at $(V,N_\infty)$ and $K$ : $K$ is compact.

[L1]

$V$ is a real vector space and $N_\infty$ is a norm on it, with $N_\infty(v) = \max\{|v_j| : j<K\}$ for any admissible $K \ge 1$ (FALSE: all norms on a real vector space are equivalent, The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ , Vector space over a field, Linear subspace of a vector space, One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $\lambda u + v \in W$ for all $\lambda \in F$ and $u, v \in W$ , Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L2]

A norm induces a metric $d(v,w) = N(v-w)$ , and $|N(v)-N(w)| \le N(v-w)$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ and is Lipschitz, hence continuous, for $d_2$ clause 1, which is stated for a norm on an arbitrary real vector space; Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L3]

Open and closed sets, balls, and boundedness (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, Open ball, closed ball and sphere in a metric space, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L4]

In ZF a compact metric space is sequentially compact (In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, Countably compact, sequentially compact and limit point compact metric spaces), and a compact subset is one whose metric subspace is compact (Open cover, subcover, compact metric space, and compact subset of a metric space, Isometry, isometric embedding, and the subspace metric on a subset); the five-way equivalence 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 is not needed and is not used.

[L5]

Every convergent sequence in a metric space is Cauchy (Every convergent sequence in a metric space is Cauchy, Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences); a subsequence is indexed by a strictly increasing map, which is injective (A strictly increasing index map satisfies $n_k \ge k$ , Injection, surjection, bijection).

[L6]

Pigeonhole: there is no injection from $\sigma(p)$ into $p$ , hence none from $\mathbb{N}$ into any natural number (The pigeonhole principle on $\mathbb{N}$ claims 1 and 4, Finite, countably infinite, countable, uncountable).

[L7]

Absolute value (Absolute value in an ordered field, Basic properties of the absolute value) and the pointwise description of $e_k$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

[L8]

A compact metric space is totally bounded (A compact metric space is complete and totally bounded, and neither implication uses any choice principle, Finite $\varepsilon$ -net and totally bounded metric space).

Refutation

technique · direct

1.1

Each $e_k$ lies in $V$ , with $K = k+1$ admissible, and $N_\infty(e_k) = 1$ ; so $e_k \in K$ for every $k$ .

L1L7

1.2

For $j \ne k$ the vector $e_j - e_k$ has coordinates $1$ at $j$ , $-1$ at $k$ and $0$ elsewhere, so $N_\infty(e_j-e_k) = 1$ , that is $d(e_j,e_k) = 1$ .

L1L2L7

1.3

$K$ is bounded, since $K \subseteq B(0,2)$ : $v \in K$ gives $d(v,0) = N_\infty(v) \le 1 < 2$ .

L1L2L3

1.4

$K$ is closed in $(V,d)$ . Let $v \notin K$ , so $N_\infty(v) > 1$ , and put $r := N_\infty(v)-1 > 0$ ; if $d(w,v) < r$ then $N_\infty(w) \ge N_\infty(v) - N_\infty(v-w) > N_\infty(v) - r = 1$ , so $w \notin K$ . Hence the complement of $K$ is open.

L2L3

2.1

No subsequence of $(e_k)$ is Cauchy in $(V,d)$ : if $l \mapsto e_{n_l}$ were, with $n$ strictly increasing and hence injective, then taking the tolerance $1/2$ would give indices $l \ne l'$ with $d(e_{n_l},e_{n_{l'}}) < 1/2$ , while $n_l \ne n_{l'}$ and step 1.2 make that distance $1$ .

step 1.2L5

3.1

Hence no subsequence of $(e_k)$ converges in the metric subspace $(K, d_K)$ , a convergent sequence being Cauchy and $d_K$ being the restriction of $d$ ; so $(K,d_K)$ is not sequentially compact.

step 2.1L4L5

4.1

If $K$ were compact then $(K,d_K)$ would be a compact metric space and hence sequentially compact, contradicting step 3.1. So [A1] is false, and with steps 1.3 and 1.4 the set $K$ is closed and bounded and not compact.

step 1.3step 1.4step 3.1A1L4

5.1

The same family shows that $(K,d_K)$ is not totally bounded, which is the property the general characterisation identifies as missing. Suppose $\{y_0,\dots,y_p\} \subseteq K$ were a finite $1/2$ -net. Assigning to each $k \in \mathbb{N}$ the least $i \le p$ with $d(e_k,y_i) < 1/2$ gives a map $\mathbb{N} \to \sigma(p)$ , which cannot be injective by pigeonhole; so there are $j \ne k$ and one $i$ with $d(e_j,y_i) < 1/2$ and $d(e_k,y_i) < 1/2$ , whence $d(e_j,e_k) \le d(e_j,y_i)+d(y_i,e_k) < 1$ , contradicting step 1.2.

step 1.2L6L8∎

Remarks

What is refuted and what is not. The claim refuted is the transfer of Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ 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 to arbitrary normed spaces. No general classification of normed spaces is asserted here; in particular the classical converse, that a normed space whose closed unit ball is compact must be finite-dimensional, is not proved anywhere here.
Why the linear structure does not help. The bisection proof of Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ 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 halves one coordinate at a time and terminates because there are finitely many coordinates. On $V$ there are infinitely many, and the standard unit vectors stay a fixed distance apart no matter how far out one looks; that is exactly the failure of total boundedness in step 5.1.
The relation to the published metric-space version. FALSE: a closed and bounded subset of a metric space is compact refutes the claim for metric spaces, and its witness is $\mathbb{N}$ carrying the metric that assigns distance $1$ to distinct points — a set with no linear structure at all. The present item refutes the narrower claim about normed spaces, on a space that is a linear subspace of a function space and carries a genuine norm, so no reader can retreat to "the counterexample was not linear".
No choice principle is used. Step 4.1 quotes only the ZF implication In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, and step 5.1 quotes A compact metric space is complete and totally bounded, and neither implication uses any choice principle, also a theorem of ZF; the equivalence 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, which carries two choice hypotheses, is deliberately avoided.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 211 results over 39 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
Riesz's lemma (Wikipedia) (standard reference, not scraped)
J. Demmel, MA221 Lecture 3: Vector Norms (standard reference, not scraped)
G. Zitelli, Math 641 Functional Analysis, Part I (standard reference, not scraped)