Short answer
At present nobody knows the answer; no concrete pair
F ⊂ ℓ₂ (finite) and X (infinite–dimensional Banach)
with “F does not embed isometrically into X’’ has been produced.
Equally, nobody has been able to prove that such a counter-example does
not exist. The problem is therefore still open.
What is known
Positive results (existence of many embeddings).
• Any finite metric space embeds isometrically in some
ℓ∞ⁿ, hence every Banach space that contains arbitrarily large
finite-dimensional ℓ∞-spaces (for example the spaces
ℓ∞, L∞, C(K) with K uncountable, etc.) already contains an
isometric copy of every finite subset of every metric space, in
particular of every finite subset of ℓ₂.
• Shkarin (2004) – every finite ultrametric embeds isometrically in
every infinite-dimensional Banach space.
• Dekster (2000) – finite Euclidean simplices with prescribed edge
lengths embed in every infinite-dimensional Banach space.
• Faver et al. (2014) – the same is true for all finite metrics that
are “roundness-2” ultrametrics.
These results show that very large families of finite subsets of
ℓ₂ always embed, so a possible counter-example F would have to be
quite special.
Partial negative information (what a counter-example would have to look like). All of the following is due to J. Kilbane (arXiv 1609.08971, arXiv 1704.00319).
(i) If X is an infinite-dimensional Banach space that does not contain an isometric copy of every finite subset of ℓ₂, then X can be chosen to be separable and to contain no subspace isomorphic to ℓ₂. In fact, for every such X there is an n≤8 and an n-point subset F⊂ℓ₂⁴ that does not embed isometrically into X.
(ii) The class
𝔽 := {F ⊂ ℓ₂ finite : there is some
infinite-dimensional X with F↛X isometrically}
(if non-empty) is “small’’: up to isometry it
has cardinality 2^{ℵ₀}; more precisely, for each fixed n the set
of n-point counter-examples is meagre in the Polish space of
all n-point metric spaces of negative type.
These results restrict quite drastically where a counter-example could hide, but they do not actually produce such an example.
Very small configurations.
For 2- and 3-point subsets of ℓ₂ it is easy to embed them
isometrically into every Banach space
(one and two distances can always be realised).
No example with 4, 5, 6, 7 or 8 points is presently known.
Relation to other open problems.
The question is a metric version of the classical linear problem
“Does every infinite-dimensional Banach space contain
a 2-dimensional Hilbert subspace?’’
– known to have a negative answer linearly (consider c₀),
but still unresolved metrically.
It also interacts with longstanding questions on equilateral sets,
on the Ribe programme, and on metric characterisations of uniform
convexity/smoothness.
To summarise
• Nobody has produced a finite set of points in ℓ₂ that fails to sit
isometrically in some infinite-dimensional Banach space,
• nor has anybody proved that such a set cannot exist.
Hence the problem remains open, although extensive partial work
(noted above) shows that any counter-example would have to be quite
special and rather small.
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version: o3-2025-04-16
Status: UQ Validated
Validated: 8 months ago
Status: Failed Human Verification
Verified: 7 months ago
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