A connected F-space

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The main result Why? The construction Sources

A connected F -space

Non impeditus ab ulla scientia

K. P. Hart

Faculty EEMCS TU Delft

Oxford, 10 August, 2006: 14:30-14:55

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Outline

1 The main result

2 _Why?

d -independent sets and d -bases What does our space do then?

3 The construction

Intuition Starting point

Thin out Su

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A space and a function

There is a compact Hausdorff space, X , that is connected and an F -space.

It supports a continuous real-valued function, f , that is not essentially constant.

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A space and a function

There is a compact Hausdorff space, X , that is connected and an F -space.

It supports a continuous real-valued function, f , that is not essentially constant.

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Contrasting behaviour of functions

For every continuous function g : X → R and every t in the interior of the interval g [X ] the interior of g←(t) is nonempty. (Follows from connected plus F .)

Yet, for f we have: Ωf =Stint f←(t) is not dense.

(This is not essentially constant.)

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Contrasting behaviour of functions

For every continuous function g : X → R and every t in the interior of the interval g [X ] the interior of g←(t) is nonempty. (Follows from connected plus F .)

Yet, for f we have: Ωf =S_tint f←(t) is not dense.

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Outline

1 The main result

2 _Why?

d -independent sets and d -bases

What does our space do then?

3 The construction Intuition Starting point Thin out Su Create X 4 Sources

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d -independent sets

D, a subset of C (X ), is d -independent if for every nonempty open set O the nonzero elements in {d O : d ∈ D} are linearly

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d -bases

A d -independent set D is a d -basis if for every g ∈ C (X ) there is a disjoint family O of open sets, with dense union, such that for every O the restriction g O is a linear combination of (finitely many members of) {d O : d ∈ D}.

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Outline

1 The main result

2 _Why?

d -independent sets and d -bases

What does our space do then?

3 The construction

Intuition Starting point Thin out Su Create X

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Maximally independent does not mean base

The family {1} is maximally d -independent.

(For every continuous function g : X → R and every t in the interior of the interval g [X ] the interior of g←(t) is

nonempty.)

Yet, the family {1} is not a d -basis.

(For f we have: Ωf =

S

tint f

←_{(t) is not dense.)}

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Maximally independent does not mean base

The family {1} is maximally d -independent.

(For every continuous function g : X → R and every t in the interior of the interval g [X ] the interior of g←(t) is

nonempty.)

Yet, the family {1} is not a d -basis.

(For f we have: Ωf =

S

tint f

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No (easy) projection

Using a d -basis that contains 1 one can project C (X ) onto the subspace of essentially constant functions, in case X is extremally disconnected.

Unknown (but wanted) for basically disconnected spaces. Apparently even more difficult for F -spaces.

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No (easy) projection

Unknown (but wanted) for basically disconnected spaces.

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No (easy) projection

Unknown (but wanted) for basically disconnected spaces. Apparently even more difficult for F -spaces.

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The main result Why? The construction Sources Intuition Starting point Thin out Su Create X

Outline

1 The main result

2 _Why?

d -independent sets and d -bases What does our space do then? 3 The construction

Intuition

Starting point Thin out Su Create X

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A rough picture

Think of X as the following subspace of S : [0, 1] × {0} ∪ C × [0, 1] (C is the Cantor set)

Think of f as resulting from the map from C onto [0, 1] and constant on complementary intervals in bottom line.

This ‘X ’ is not an F -space . . .

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A rough picture

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A rough picture

This ‘X ’ is not an F -space . . .

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Outline

1 The main result

2 _Why?

Intuition

Starting point

Thin out Su Create X

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A particular β

Let S be the unit square [0, 1]2

Let S = ω × S

Define p : S → [0, 1] by p(n, x, y ) = x and extend to βp : βS → [0, 1].

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A particular β

Let S = ω × S

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A particular β

Let S = ω × S

Define p : S → [0, 1] by p(n, x, y ) = x

and extend to βp : βS → [0, 1].

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A particular β

Let S = ω × S

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A component of βS and a function

βπ : βS → βω is the extension of π : hn, x, y i 7→ n.

Pick one u ∈ βω \ ω.

Let Su= βπ←(u)

Su is a compact connected F -space

βp Su is continuous

but Su and βp are not good enough . . .

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A component of βS and a function

βπ : βS → βω is the extension of π : hn, x, y i 7→ n. Pick one u ∈ βω \ ω.

Let Su= βπ←(u)

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A component of βS and a function

Let Su= βπ←(u)

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A component of βS and a function

Let Su= βπ←(u)

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A component of βS and a function

Let Su= βπ←(u)

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A component of βS and a function

Let Su= βπ←(u)

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Outline

1 The main result

2 _Why?

d -independent sets and d -bases What does our space do then? 3 The construction Intuition Starting point Thin out Su Create X 4 Sources

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Get rid of interiors

Set Y0 = Su and q0 = βp Y0 and recursively

Yα+1 = Yα\Stintαqα←(t) and qα+1 = qα Yα+1

(intα: interior in Yα)

Yα =Tβ<αYβ and qα = q0 Yα if α is a limit

There is a first (limit) δ < c+ where Yδ= Yδ+1, meaning that

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Get rid of interiors

intδq_δ←(t) = ∅ for all t

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Get rid of interiors

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Get rid of interiors

intδq←_δ (t) = ∅ for all t

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Outline

1 The main result

2 _Why?

Intuition Starting point Thin out Su

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Tie everything together

Sadly, Yδ is not connected

However, take the bottom line of Su:

Bu= Su∩ cl ω × [0, 1] × {0}.

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Tie everything together

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Tie everything together

Bu= Su∩ cl ω × [0, 1] × {0}.

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Here are X and f

Finally then

X = Bu∪ Yδ

f = βp X X is connected and F int f←(t) ⊆ Bu for all t

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Here are X and f

Finally then

X = Bu∪ Yδ

All components of Yδ meet the top line, so Ωf ⊆ Bu is not dense

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Here are X and f

Finally then

X = Bu∪ Yδ

f = βp X

X is connected and F int f←(t) ⊆ Bu for all t

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Here are X and f

Finally then

X = Bu∪ Yδ

f = βp X X is connected and F

int f←(t) ⊆ Bu for all t

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Here are X and f

Finally then

X = Bu∪ Yδ

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Here are X and f

Finally then

X = Bu∪ Yδ

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Light reading

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

Y. A. Abramovich and A. K. Kitover.

d -Independence and d -bases, Positivity, 7 (2003), 95–97.

K. P. Hart.