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A Lambda Calculus for Real Analysis

Paul Taylor 1 Andrej Bauer 2

1Department of Computer ScienceUniversity of Manchester

UK EPSRC GR / S58522

2Department of Mathematics and PhysicsUniversity of Ljubljana

Computability and Complexity in AnalysisSunday, 28 August 2005

www.cs.man.ac.uk / pt / ASD
http://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://www.cs.man.ac.uk/~pt/ASD/index.pdfhttp://find/

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All functions are continuous and computableThis is not a Theorem ( a la Brouwer) but a design principle.The language only introduces continuous computablefunctions.
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In particular, all functions R R are continuousand correspond to open subspaces.
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Hence a < b, a > b and a b are denable, but a b, a b and a = b are not denable.
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This is because R is Hausdor ff but not discrete.
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This is because R is Hausdor ff but not discrete.

N and Q are discrete and Hausdor ff .

So we have all six relations for them.

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Geometric, not Intuitionistic, logicA term : is called a proposition .A term : X is called a predicate or open subspace.

We can form and .Also n : N . x, q : Q . x, x : R . x and x : [0, 1]. x.
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Geometric, not Intuitionistic, logicA term : is called a proposition.A term : X is called a predicate or open subspace.

We can form and .Also n : N . x, q : Q . x, x : R . x and x : [0, 1]. x.But not x : X . x for arbitrary X it must be overt .
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Geometric, not Intuitionistic, logicA term : is called a proposition.A term : X is called a predicate or open subspace.

We can form and .Also n : N . x, q : Q . x, x : R . x and x : [0, 1]. x.But not x : X . x for arbitrary X it must be overt.

Negation and implication are not allowed.Because:

this is the logic of open subspaces ;

the function on

is not continuous ;

the Halting Problem is not solvable.
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Universal quanticationWhen K X is compact (e.g. [0, 1] R ), we can form x : K . x.

x : K x================= x : K . x
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x : K x================= x : K . x

The quantier is a (higher-type) function K : K .

Like everything else, its Scott continuous .

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x : K x================= x : K . x

The quantier is a (higher-type) function K : K .

Like everything else, its Scott continuous .The useful cases of this in real analysis are

x : K . > 0.(x, ) > 0.x : K .(x, )

x : K .n.(x, n) n.x : K .(x, n)in the case where ( 1 < 2) (x, 2) (x, 1)or (n1 > n2) (x, n2) (x, n1).

Recall that uniform convergence, continuity, etc.involve commuting quantiers like this.

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Propositions and statementsWhats the problem with ?
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Propositions and statementsWhats the problem with ? We cant write > 0!
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Propositions may be computationally observable.

Equations and implications amongst propositions or predicatesmay be logically provable from the axioms.We call them statements.
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For example, with a ( d. d < a), a = b iff a = b and a b iff a bin the arithmetical order.Hence a b, a b and a = b are meaningful,as statements, not as propositions .
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For example, with a ( d. d < a), a = b iff a = b and a b iff a bin the arithmetical order.Hence a b, a b and a = b are meaningful,as statements, not as propositions.

In fact, a b is equivalent as a statement to (a > b) ,and a = b to (a b) .

We deal with > 0 by allowing as parameter or free variable .
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Examples: continuity and uniform continuityRecall that, from local compactness of R ,

x > 0. y: [x ]. y

Theorem: Every denable function f : R R is continuous:

> 0 > 0. y: [x ]. fy fx <

Proof: Put x, y fy fx < , with parameters x, : R .

Theorem: Every function f is uniformly continuouson any compact subspace K R :

> 0 > 0. x : K . y : [x ]. fy fx <

Proof: > 0 and x : K commute.
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Example: Dinis theoremTheorem: Let f n : K R be an increasing sequence of functions

n : N , x : K f nx f n+ 1x : Rthat converges pointwise to g : K R , so

> 0, x : K n. gx f nx < .

If K is compact then f n converges to g uniformly .

Proof: Using the introduction and Scott continuity rules for ,

> 0 x : K . n. gx f nx < n. x : K . gx f nx <

Corollary: Since ASD has a computational interpretation,Dinis theorem is computationally valid.

l f b
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Relative containment of open subspacesLet , , be propositions with parameters x1 : X 1, ..., xk : X k .(We conventionally write for this list.Semantically, is the space X 1 X k .)Then , , dene open subspaces of .They satisfy a Gentzen-stle rule of inference:

, ====================

in which the top line means

within the open subspace of dened by ,

the open subspace dened by is contained in the open subspace dened by .

and the bottom line meansthe intersection of the open subspaces dened by and

is contained in that dened by .

R l i i f l d b
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Relative containment of closed subspacesLet , , be propositions with parameters x1 : X 1, ..., xk : X k .(We conventionally write for this list.Semantically, is the space X 1 X k .)Then , , dene closed subspaces of .They satisfy a Gentzen-stle rule of inference:

, ====================

in which the top line means

within the closed subspace of dened by ,

the closed subspace dened by contains the closed subspace dened by .

and the bottom line meansthe intersection of the closed subspaces dened by and

is contained in that dened by .

E i f !
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Exercise for everyone!Make a habit of trying to formulate statements in analysisaccording to (the restrictions of) the ASD language.

This may be easy it may not be possible

The exercise of doing so may be 95% of solving your problem!

C t ti i t di t l th
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Constructive intermediate value theoremSuppose that f : R R doesnt hover , i.e.

b, d : R b < d x. (b < x < d) ( fx

0),and f 0 < 0 < f 1. Then fc = 0 for some 0 < c < 1.

Interval trisection : Let a0 0, e0 1,

bn 13(2an + en) and dn 13 (an + 2en).

Then f (cn) 0 for some bn < cn < dn , so put

an

+1, e

n+

1 an , cn if f (cn) > 0

cn , en if f (cn) < 0.

Then f (an) < 0 < f (en) and an c en .(This isnt the ASD proof / algorithm yet!)

St bl
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Stable zeroesThe interval trisection nds zeroes with this property:

fd fba b c d e

d e a b c

fb fd

Denition: c : R is a stable zero of f if

a, e : R a < c < e bd. (a < b < c < d < e)

( fb < 0 < fd fb > 0 > fd).

The subspace Z [0, 1] of all zeroes is compact .The subspace S [0, 1] of stable zeroes is overt (as we shallsee...)

Straddling intervals
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Straddling intervalsAn open subspace U R contains a stable zero c U S iff U also contains a straddling interval ,

[b, d] U with fb < 0 < fd or fb > 0 > fd.

[ ] From the denitions. [ ] The straddling intervalis an intermediate value problem in miniature.

Straddling intervals
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Straddling intervalsAn open subspace U R contains a stable zero c U S iff U also contains a straddling interval ,

[b, d] U with fb < 0 < fd or fb > 0 > fd.

[ ] From the denitions. [ ] The straddling intervalis an intermediate value problem in miniature.

Notation: Write U if U contains a straddling interval.We write this containment in ASD using the universalquantier.

bd. (x : [b, d]. x) ( fb < 0 < fd) ( fb > 0 > fd).

The possibility operator

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By hypothesis, (0, 1) , whilst trivially.

iI U i i. U i.If f : R R is an open map , this is easy.

If f : R R doesnt hover, it depends on connectedness of R .


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By hypothesis, (0, 1) , whilst trivially.

iI U i i. U i.If f : R R is an open map , this is easy.

If f : R R doesnt hover, it depends on connectedness of R .

Denition: A term : X

with this propertyis called an overt subspace of X .

A simpler example: For any point a : X ,the neighbourhood lter a . a is a possibilityoperator.

is a point iff it also preserves and .

The Possibility Operator as a Program

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The Possibility Operator as a ProgramTheorem: Let be an overt subspace of R with .Then has an accumulation point c R ,

i.e. one of which every open neighbourhood c U Rsatises U :

: R c

Example: In the intermediate value theorem, any such c is a

stable zero.Proof: Interval trisection.

Corollary: Obtain a Cauchy sequence from a Dedekind cut .

(I expect to get a representation 2 N

R in the sense of TTE by proving a result of Brattko & Hertling in ASD.)

Possibility operators classically

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Possibility operators classicallyDene U as U S , for any subset S X whatever.

Then iI U i iff i. U i.

Conversely, if has this property, let

S {a X | for all open U X , a U U }

W X \ S = {U open | U }

Then W is open and S is closed. W by preservation of unions.Hence U holds iff U W , i.e. U S .

If had been derived from some S then S = S , its closure.Classically, every (sub)space S is overt.

Necessity operators
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Necessity operatorsLet K R be any compact subspace.(For example, all zeroes in a bounded interval.)

U (K U ) is Scott continuous .

Notation: Write for x : K . x.

Modal operators separately
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Modal operators, separatelyencodes the compact subspace Z {x I | fx = 0}of all zeroes.

encodes the overt subspace S of stable zeroes.

X is true and U V (U V )

is false and (U V ) U V .

(Z ) iff is false(S ) iff R is true

Modal operators, together
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Modal operators, together and for the subspaces S Z are related in general by:

U

V

(U

V )U (U W = X )

V (V W )

S is dense in Z iff

(U V ) U V

V (V W )

In the intermediate value theoremfor functions that dont hover ( e.g. polynomials):

S = Z in the non-singular case S Z in the singular case (e.g. double zeroes).

Modal laws in ASD notation
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Modal laws in ASD notationOvert subspace Compact subspace

( ) ( )

( ) ( x. x)Commutative laws:

x. ( y. xy) y. ( x. xy)

x. ( y. xy) y. ( x. xy)

Mixed modal laws for a compact overt subspace.

( ) and ( )

Empty / inhabited is decidable
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p yTheorem: Any compact overt subspace ( , ) is either empty( ) or non-empty ( ).

Proof: empty inhabited complementary

( ) (mixed) ( )

Empty / inhabited is decidable
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p yTheorem: Any compact overt subspace ( , ) is either empty( ) or non-empty ( ).

Proof: empty inhabited complementary

( ) (mixed) ( )

The dichotomy (either or ) means thatthe parameter space is a disjoint union .

So, if it is connected , like R n ,

something must break at singularities .It is the modal law ( ) .

Non-empty compact overt subspace of R has a

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p y p pmaximum

Theorem: Any overt compact subspace K R is

either empty or has a greatest element, max K K .

Denition: max K satises, for x : R ,

(x < max K ) (k : K . x < k )(max K < x) (k : K . k < x)

k : K k max K

, k : K k x max K x

Compact overt subspace of R has a maximum
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p pProof: Dene a Dedekind cut (next slide)

d k : K . d < k and u k : K . k < u

Hence there is some a : R with

d (d < a) and u (a < u)

Moreover, a K .K is also the closed subspaceco-classied by x ( k . x k ),so we must show that a .

a ( k . a k ) ( k . a < k ) (k < a) ( k . a < k ) ( k . k < a) a a (a < a) (a < a) .

Compact overt subspace of R denes a Dedekind cut
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Overt subspace Compact subspace

, , and so R commutes with , and

d ( k . d < k ) Dedekind cut u ( k . k < u)

(d < e) e lower / upper t (t < u) (d < e) ( k . e < k ) ( k . k < t) (t < u) ( k . d < e < k ) (Frobenius / ) ( k . k < t < u) ( k . d < k ) d (transitivity) ( k . k < u) u

rounded (interpolation)

d. d d. ( k . d < k ) inhabited u. u u. ( k . k < u) ( k . d. d < k ) (directed joins) ( k . u. k < u)

(inhabited) (extrapolation)

The Bishop-style proof
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Denition: K is totally bounded if, for each > 0,theres a nite subset S K such that

x : K . y S . x y < .Proof: If K is closed and totally bounded, either the set S1 is empty, in which case K is empty too, or xn max S2 n denes a Cauchy sequencethat converges to max K .

But K is also overt , with > 0. y S . y.

Denition: K is located if, for each x X ,inf {|x k | | k K } is dened.(A diff erent usage of the word located.)closed, totally bounded compact and overt located(in TTE) also r.e. closed

Total boundedness and locatedness are metrical concepts.

Compactness and overtness are topological.

The real interval is connected (usual proof)
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Any closed subspace of a compact space is compact.Any open subspace of an overt space is overt.

Any clopen subspace of an overt compact space is overtcompact, so its either empty or has a maximum.

Since the clopen subspace is open, its elements are interior,so the maximum can only be the right endpoint of the interval.

Any clopen subspace has a clopen complement. They cant both be empty, but in the interval they cant both have maxima (the rightendpoint).

Hence one is empty and the other is the whole interval.

Connectedness in modal notation

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We have just proved

( ) , ( )

where x : [0, 1]. x and x : [0, 1]. x.

Using the mixed modal law ( )and the Gentzen-style rules

=================

=================

connectedness may be expressed in other ways:

( ) ( ) ( ) ( )

( ) ( )( ) ( )

Weak intermediate value theorems
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Let f : [0, 1] R , and use two of these forms of connectedness.

Put x (0 < fx) and x ( fx < 0).Use ( ) = ( ) . ( ) by disjointness.Then ( f 0 < 0 < f 1) x : [0, 1]. fx 0 .So the closed, compact subspace Z {x : I | fx = 0} is not empty.

Put x (e < fx) and x ( fx < t).Use ( ) ( ).

( ) by locatedness.Then ( f 0 < e < t < f 1) x : [0, 1]. e < fx < t .or > 0 x. fx < .So the open, overt subspace {x | e < fx < t} is inhabited.

Straddling intervals in ASD
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Let f : [0, 1] R be a function that doesnt hover.

Proposition: preserves joins, (n. n) n. n .Proof: Consider x n. y. (x < y < u) ( fy >< 0) z : [x, y]. n z.Then x. + x x by connectness.

Lemma: 0 < a < 1 is a stable zero of f iff it is an accumulation point of , i.e. a .

Theorem: and obey ( ).

They also obey ( ) iff f doesnt touch the axis without crossing it.When f is a polynomial, this is the non-singular case, where f has no zeroes of even multiplicity.

Solving equations in ASD
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In the non-singular case, all zeroes are stable, and dene a non-empty overt compact subspace,

which has a maximum .So the classical textbook proof of IVT,

a sup {x : [0, 1] | fx 0},

is computationally meaningful!The set of zeroes varies discontinuously at singularities in theparameters.The modal operators and are Scott-continuousthroughout the parameter space.The interval trisection algorithm for nds some zero,even in the singular case, but it behaves non-deterministically and catastrophically.

Diff erentiation
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Dene ( fx, f x) together by a Dedekind cross-hair .

Characterise ( e0 < fx < t0) (e1 < f x < t1) by

. h : [0, ]. e1 + e1h < f (x + h) < t0 + t1h e1 t1h < f (x h) < t0 e1h

This is a Dedekind cut in ( e0, t0) since f :R R

is a function.It is bounded in (e1, t1) if f is Lipschitz at x.

It is a Dedekind cut in (e1, t1) if f is diff erentiable at x.

I need help!

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Im a categorist, not an analyst.

I last did real analysis as a second year undergraduate.

I need a real analyst to set an agenda for me.

I also need a job from September 2006.
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