There is no categorical metric continuum

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The main result The Proof Sources

There is no categorical metric continuum

Non impeditus ab ulla scientia

K. P. Hart Faculty EEMCS

TU Delft

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Outline

1 The main result

The statement What it means 2 The Proof Main lemma Finishing up 3 Sources

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The main result The Proof Sources The statement What it means

Outline

1 The main result

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The statement

Given a metric continuum X there is another metric continuum Y such that

X and Y look the same

(they have elementarily equivalent countable bases) X and Y are not homeomorphic

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The statement

(they have elementarily equivalent countable bases)

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The statement

(they have elementarily equivalent countable bases) X and Y are not homeomorphic

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Outline

1 The main result

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Elementary equivalence

We consider bases that are closed under finite unions and intersections.

These are lattices and ‘elementary equivalence’ is with respect to the lattice structure.

Two lattices are ‘elementarily equivalent’ if they satisfy the same first-order sentences.

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Example: zero-dimensionality

Here is a first-order sentence, call it ζ (∀x )(∀y )(∃u)(∃v )

(x u y = 0) → ((x 6 u) ∧ (y 6 v ) ∧ (u u v = 0) ∧ (u t v = 1))

In words: any two disjoint closed sets (x and y ) can be separated by clopen sets (u and v ).

By compactness, if some base satisfies this sentence then the space is zero-dimensional.

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Example: zero-dimensionality

(x u y = 0) → ((x 6 u) ∧ (y 6 v ) ∧ (u u v = 0) ∧ (u t v = 1))

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Example: zero-dimensionality

(x u y = 0) → ((x 6 u) ∧ (y 6 v ) ∧ (u u v = 0) ∧ (u t v = 1))

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The statement What it means

Example: no isolated points

Here is a another first-order sentence, call it π

(∀x )(∃y ) (x < 1) → ((x < y ) ∧ (y < 1))

In words: every closed proper subset (x ) is properly contained in a closed proper subset (y );

in fewer words: there are no isolated points.

If some base satisfies this sentence then the space has no isolated points.

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Example: no isolated points

(∀x )(∃y ) (x < 1) → ((x < y ) ∧ (y < 1))

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Example: no isolated points

(∀x )(∃y ) (x < 1) → ((x < y ) ∧ (y < 1))

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Example: no isolated points

(∀x )(∃y ) (x < 1) → ((x < y ) ∧ (y < 1))

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Example: the Cantor set is categorical

Let X be compact metric with a countable base B for the closed sets that satisfies ζ and π.

Then X is zero-dimensional and without isolated points. So X is (homeomorphic to) the Cantor set C .

Thus: if X looks like C then X is homeomorphic to C .

The Cantor set iscategoricalamong compact metric spaces.

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Example: the Cantor set is categorical

Then X is zero-dimensional and without isolated points.

So X is (homeomorphic to) the Cantor set C .

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Example: the Cantor set is categorical

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Example: the Cantor set is categorical

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Example: the Cantor set is categorical

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What the main result says

Among metric continua there is nocategoricalspace.

No (in)finite list of first-order properties will characterize a single metric continuum.

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What the main result says

Among metric continua there is nocategoricalspace.

No (in)finite list of first-order properties will characterize a single metric continuum.

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A case in point: the pseudoarc

Thepseudoarc is the only metric continuum that is

hereditarily indecomposable and chainable

A two-item list but . . . Chainability is not first-order. (Hereditary indecomposability is.)

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A case in point: the pseudoarc

Thepseudoarc is the only metric continuum that is hereditarily indecomposable and

chainable

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A case in point: the pseudoarc

chainable

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A case in point: the pseudoarc

chainable

A two-item list but . . .

Chainability is not first-order. (Hereditary indecomposability is.)

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A case in point: the pseudoarc

chainable

A two-item list but . . . Chainability is not first-order.

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A case in point: the pseudoarc

chainable

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The main result The Proof Sources Main lemma Finishing up

Outline

1 The main result The statement What it means 2 The Proof Main lemma Finishing up 3 Sources

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An embedding lemma

Let X and Z be metric continua, with countable lattice bases, B and C, for their respective families of closed sets.

Let u be a free ultrafilter on ω.

There is an embedding of C into the ultrapower of B by u.

Ultrapower Bu: the power Bω modulo the equivalence relation

“{n : Bn= Cn} ∈ u”.

Ultrapower theorem: B and Bu are elementarily equivalent.

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An embedding lemma

“{n : Bn= Cn} ∈ u”.

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An embedding lemma

“{n : Bn= Cn} ∈ u”.

Ultrapower theorem: B and Bu are elementarily equivalent.

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How to make Y

Let ϕ : C → Bu be an embedding.

Apply the L¨owenheim-Skolem theorem:

Find a countable elementary sublattice D of Bu that contains ϕ[C].

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How to make Y

Let Y be the Wallman space of D.

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How to make Y

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How to make Y

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How to make Y

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How to make Y

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Properties of Y

Y is compact metric (D is countable).

D is a base for the closed sets of Y (by Wallman’s theorem).

D is elementarily equivalent to Bu and hence to B.

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Properties of Y

Y maps onto Z (because ϕ[C] is embedded into D).

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Properties of Y

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Properties of Y

Y maps onto Z (because ϕ[C] is embedded into D).

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Outline

1 The main result The statement What it means 2 The Proof Main lemma Finishing up 3 Sources

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Getting a good Y

Let X be given, with a countable base B for its closed sets.

There is a metric continuum Z that is not a continuous image of X (Waraszkiewicz).

Find Y with a base that is elementarily equivalent to B and such that Y maps onto Z .

So: Y is not homeomorphic to X .

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Getting a good Y

Let X be given, with a countable base B for its closed sets. There is a metric continuum Z that is not a continuous image of X (Waraszkiewicz).

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Getting a good Y

Find Y with a base that is elementarily equivalent to B and

such that Y maps onto Z .

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Getting a good Y

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Getting a good Y

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Light reading

Website: fa.its.tudelft.nl/~hart

T. Banakh, P. Bankston, B. Raines and W. Ruitenburg.

Chainability and Hemmingsen’s theorem,

http://www.mscs.mu.edu/~paulb/Paper/chainable.pdf

K. P. Hart.

There is no categorical metric continuum, to appear.

Z. Waraszkiewicz.

Sur un probl`eme de M. H. Hahn, Fundamenta Mathematicae