CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Identity is compact iff the space is finite dimensional

Statement refuted

The false general statement is: the identity operator of every normed space is compact. In fact, for a normed space $X$ over $R$ or $C$ the identity $I_{X}$ is compact (Compact linear operator) if and only if $X$ admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and the identity of the infinite-dimensional space $ℓ^{2} (N, K)$ is not compact (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

Facts & Assumptions

[A1]

$I_{X}$ is compact exactly when $\overline{I_{X} ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator, Open ball, closed ball and sphere in a metric space); the closed unit ball is closed, so $\overline{{\overline{B}}_{X}} = {\overline{B}}_{X}$ (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[A2]

${\overline{B}}_{X}$ is compact if and only if $X$ admits an ordered basis of finite length (The closed unit ball is compact if and only if the normed space is finite-dimensional).

[A3]

In a vector space with a spanning subset of size $n$ , every linearly independent subset is finite of size at most $n$ (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ , Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[A4]

In $ℓ^{2} (N, K)$ the standard vectors $e_{k}$ satisfy $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ and $∥ e_{k} ∥_{2} = 1$ , and the pairing is $⟨ a, b ⟩ = \sum_{n} a_{n} \overline{b_{n}}$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

Counterexample

technique · direct

Given: A normed space $X$ over $R$ or $C$ , and the sequence space $ℓ^{2} (N, K)$ with its standard vectors $e_{k}$ .

1.1A1

$I_{X}$ is compact if and only if ${\overline{B}}_{X}$ is compact, because $I_{X} ({\overline{B}}_{X}) = {\overline{B}}_{X}$ is already closed.

1.2A4

In $ℓ^{2} (N, K)$ the vectors $e_{0}, \dots, e_{n}$ are linearly independent for every $n$ : if $\sum_{k \leq n} c_{k} e_{k} = 0$ , then pairing with $e_{j}$ gives $c_{j} = ⟨ \sum_{k} c_{k} e_{k}, e_{j} ⟩ = 0$ for every $j \leq n$ .

2.1step 1.1A2

Hence $I_{X}$ is compact if and only if $X$ admits an ordered basis of finite length, by [step 1.1] and [A2].

2.2step 1.2A3

The space $ℓ^{2} (N, K)$ admits no ordered basis of finite length: if it admitted a spanning list of length $n$ , then by [A3] every linearly independent subset would have at most $n$ elements, contradicting the independent list $e_{0}, \dots, e_{n}$ of [step 1.2] of length $n + 1$ .

3.1step 2.1step 2.2∎

Therefore the identity of $ℓ^{2} (N, K)$ is not compact, by [step 2.1] and [step 2.2]; this is the promised witness, and [step 2.1] is the asserted equivalence.

Depends on

Used by

Compactness is not preserved by strong operator limits Counterexample

Dependency tree · two levels

72 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

Theo Bühler and Dietmar Salamon, Functional Analysis — §4.2 p.184, the identity is compact only in finite dimension (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.1, examples and counterexamples for compactness (standard reference, not scraped)