CHAPTER III

COMPARISON OF COVERING CLASSES

§ 1. Doubly-starred Sets

Given any set A in 1

2, define the transpose of A, tr A, by tr A = ((x, y):(y, x) EA)

If (a, b) is a point of 1

2 and a_;:: b, write

(a,b) s = ((x,y):x_;::y, ay_;::bx, (t-b)(t-x)_;::(l-a)(l-y))

and if b > a ,

(a, b)s = tr (b, a)⁸ ••

Given any set A::; 1

2, write

As = U((a, b)s:(a, b) EA)

A set A will be called doubly-starred with respect to the points (O, 0) and (1, 1), or more simply, doubly-starred in 1

2, provided that A = As.

The next few lemmas describe some prop~rties of doubly-starred sets. Lemma 1. If A= A, then As = As •

Proof: 'Suppose (x ,y) E As with x > y (n = 1, 2, ••. ) and

n n n - n

yn ... y. Then x

_>

y. Let a

>

b be such that (x , y ) E (a , b ) , s

n - n n n n n

(a , b ) E A Since A= A, there is a subsequence (a., b .) - (a, b) E A.

n n J J

Since

a.y. > b .x. , J J - J J

(l-b.)(1-x.) > (l-a.)(1-y.) ,

J J - J J

it follows that ay .=::bx and (1-b)(l-x) .=:: (1-a)(l-y). Thus (x,y) E (a, b)s CA⁸ , whichshows A⁵ =A⁵•

If A is a set in 1

2, then the sets tA and A+ s for real s and t are defined by

tA = ((ta, tb):(a, b) EA} , A+ s = ((a+s, b+s):(a, b) EA}

Lemma 2. A = A ⁶, if and only if

tA U ( tA + 1- t) ~ A ; for all 0 < t < 1.

Proof: If A= A , 0 _:: t _:: 1, (a, b) E A with a.=:: b, then s

ta·> tb , a(tb) _=:: b(ta) ,

(1-b)(l-ta) = (1-b)(l-a) + a(l-b)(l-t)

> (1-b)(l-a) + b(l-a)(l-t) = (1-a)(l-tb) . This shows (ta, tb) E (a, b)s. Moreover, since

ta+ 1-t _=:: tb + 1-t , a( tb+ 1-t) .=:: b( ta+ 1-t) ,

(1-b)(l-(ta+l-t)) = (1-b)t(l-a) = (1-a)(l-(tb+l-t)) ,

i t follows that (ta+l-t, tb+l-t) E (a, b) . s Thus

tA

U (

tA+ 1- t) .'.:: As = A •

On the other hand, let (x, y) E (a, b) s with (a, b) E A. Without loss of generality, suppose a> b. If x = y, then ( 1, 1) EA implies (x, y) E xA c A, by hypothesis. Thus suppose x > y, and so a> b. Write

=

^(x-y)

t (x-y} + (by-ax}

and

= (x-y) + (by-ax)

s a-b

Since by.?: ax, it follows that 0 ~ t ~ 1 and s > O. Writing

(x-y) +(by-ax)= (1-a)(l-y) - (1-b)(l-x) + a - b , shows that s < 1. Since

and

s(tb+ 1-t) = s( y(a-b)

~

^{= y ,}

x-y

+by - axJ

it follows that (x, y) E s(tA+ 1-t). Since s(tA+ 1. t) ~ A, it follows that

(x, y) EA. s s

Thus (a, b) ~A which shows that A c: A .

Lemma 3. (As )⁶ = A⁶ •

Proof: Suppose (a, b) E A with a> b and assume (x, y) E (a, b) • Since, for 0 s ~ t ~ 1,

tx ~ ty ' aty

>

btx ,

(1-b)(l-tx) = (1-b)(l-x) + x(l-t)(l-b)

> ( 1 - a)( 1 -y) + y( 1- t)( 1 - a) = ( 1 - a)( 1- ty)

it follows that (tx, ty) E (a, b) s. Moreover,

tx+ 1-t

>

ty+ 1 - t '

a(ty+l-t) = aty + a(l-t) ~ btx + b(l-t) = b(tx+l-t) , (1-b)(l-(tx+l-t)) = (1-b)(l-x)t

> (

1 - a)( 1-y) t = ( 1 - a)( 1 -( ty+ 1 - t) ) , . and so (tx+l-t, ty+l-t) E (a, b) • s It follows that

t( a, b) s U ( t( a, b) s + 1-t)

=.

^{(a, b) 5 ((a, b) E A)}

for all t, 0 < t < 1. This implies

tA s U ( tA s + 1 - t)

=-

^{As ,}

for 0

<

t

<

l; and s·o by Lemma 2, (As}s = As.

The next lemma asserts that the transpose of a doubly- starred set is again doubly-starred.

Lemma 4. s s

If A= A , then trA =(tr A) Proof: tr(As) = tr(U ((a, b)s:(a, b) EA})

= U (tr(a, b)s:(a, b) EA}

= U((b,a)s:(a,b) EA}

=

^{U ((b, a) s: (b, a) E tr A}}

=

^{(tr A)} s ^•

s . s

Since A= A , it follows that trA= (trA).

Under the same assumption made in Lemma 4, it can be

- - s

shown that A = (A) •

Lemma 5. s - - s

A= A implies A= (A) •

Proof: By Lemma 2, tA c A. Since tA = tA, it follows that tA c A. Similarly

tA + 1-t = (tA+l-t) c A

- - s

so that by Lemma 2, A = (A) . Given any set A~ 1

2, and t, 0 < t < 1, define a function A(t) by

A(t) = inf(y:(t, y) E A)

where A(T) is defined to be equal 1, if (y:(t, y) E A) = ¢.

Lemma 6. s

H A= A , then the functions A(t) and trA(t) ~

continuous on (0, 1), non-decreasing on [O, l] , and

( 3. 1) A( t)

It ,

tr A( t) It , ( l -A( t))

I (

l - t) , ( 1 -tr A( t))

I (

1- t) , are all non-decreasing on (O, l).

Proof: Given 0 < t < 1, it follows that (t, A(t)) E A. By

- - s

Lemma 5, A= (A) , so that for 0 <a.:::_ 1, (a.t, a.A(t)) EA and

(at+l-a, aA(t)+l-a.) EA. Then

A(a. t) _:: a A( t) and

A(a t+ 1-a.) _:: a.A( t) + 1-a. • These inequalities imply

and

A(a. t) < A( t)

ctt

^t

1 - A(a.t+l-a.) < 1 - A(t) 1 - (at+ 1-a) - 1 - ^{t '}

which shows that A(t)

It

and ( 1-A(t))

I(

1-t) are non-decreasing on (O, 1).

Since tr A = (tr A) s, it follows that tr A(t)

It

and ( 1-tr A(t))

I(

1-t} are non- decreasing on (0, 1). If 0 < x < y, then

A(x) < ~ A(y) < A(y) , - y

so that A(t) is non-decreasing. The same is true for trA(t). Since A= As implies (O, O) EA and (x, x) EA, it follows that A(O) = 0 and A(x) _:: x, which shows A(x) is continuous at x = O. If t > 0 and 0

<

x

<

1,

so that

1 - A(x+t) 1 - A(x)

l - (

x+ t) >

l -

x

A(x} .::_ A(x+t) _:: l :x + ( 1 -: l

:x)

^{A(x) .}

By the same reasoning,

A(x)

>

A(x-t)

~

( 1 + l:x)

A(~)

^{- l:x ·}

These relations show that A(x) is continuous at each 0

<

x

<

1, and the lemma is proved.

§ 2. Main Results

In Chapter II, the partial ordering on covering classes:

d8(£)

.:S.

dT(f) for all f E J, was introduced and studied. The question

arises as to what can be said in general about the behavior of d

8(£) and dT(f) when the covering classes S and T are not necessarily com-

parable in this ordering. For this purpose the set

is now studied.

The main results concerning the dimension of functions state that the set R(S, T) is closed and doubly-st'3;rred in 1

2; and conversely, any closed, doubly..: starred set in 1

2 is of the form R(S, T) for some covering classes S and T. Apart from the inherent interest in these

results, they take on significance when, in Chapter IV it is shown that they have direct application to the study of the dimension of closed sets.

Theorem 1. R(S, T) s

= (R(S,T)).

Proof: Suppose (a., 13) E R(S, T). Then there is f E J with

t t 1-t

If 0 < t < 1, then both f(x) and f(x) · x are in J by Theorem 3 (§2,I). Since

and

for every coyering class A, it follows that

(ta.,

t13) E R(S, T) and (ta+ 1-t, tl3 + 1-t) E R(S, T). By Lemma 2, § 1, it follows that R(S, T)

=R(S,T). s

The set R(S, T) can be described in terms of the special functions introduced in §3, Chapter II.

Lemma. R(S, T) = ((a, 13):a;:: 13;:: f(a) or 13 >a;:: g(a)} , where (a., 13) E 12 and,

(3. 2) and ( 3. 3)

f( 1) = sup f(a.) a.< 1

g( 1) = sup g(a.) a.< 1 Proof: If f E J and d

8^(£) =a. , a. < 1, then

by Theorem 2 ( § 3, II). Further, by Theorem 3 ( § 3, II), f E J and d8(f) = 1 implies dT{f) ;:: sup dT( (xa.)

8) , for any covering classes S a.< 1

and T. Consequently

R(S, T) ,:: ((a., 13) :a.;:: 13 ;:: f(a.) or 13 ;:: a;:: g(13)} • On the other hand, suppose

Then put f = (xa.)

5 /\ xl3. By Lemma 2 (§2,II)

and

Now, if ex.

=

^{1 and 1 _?: 13 _?: sup}

1

dT( (xcx.)S) , let g be the function defined ex.<

in the proof of Theorem 3, ( § 3, II) such that d

5(g)

=

1 and dT(g)

= sup dT( (xcx.)

5). Write h = g /\xi' and, as before, it follows that

This shows that

d5(h)

=

¹

dT(h) = 13

[(ex., 13) :ex._?: 13 _?: f(cx.) or 13 _?:ex.

>

g(cx.)} ~ R(S, T) and the proof is complete.

Theorem 2. R(S, T) is closed in 1 2.

Proof: Writing R = R(S, T), the lemma above implies that R(t)

=

^f(t)

tr R ( t) = g ( t) ,

where f and g are defined by (3. 2) and (3. 3). By .Lemma 6, §1, f and g are continuous on [O, 1

J

so that R(S, T) is closed.

The values that (d

5(f), dT(F)) may take, in general, are quite unrestricted as the following theorem demonstrates.

Theorem 3.

..!!_

R is any closed, doubly-starred set in 1 2, there are covering classes S and T such that R.

=

^{R(S, T).}

Proof: For each positive integer k, define

and

gk(x)

= Ti<

^{V trR(x) .}

Then the functions fk and gk decrease monotonically to R(t) and trR(t) respectively; and they are continuous on [0, l], since R(t) and trR(t) are continuous. Further, since

and

gk(x) = 1 V trR(x)

x

"Zk

^x

=

=

1 - x/2k

----

_{1 - x} ^/\

1 - x/2k

1 -

x ^/\

1- R(x)

l -

x

1 - trR(x) 1 - x

the functi.ons 1 - x/Zk b ·

l _

x eing non- ecreasing in x, it is c ear d · · · · 1 th at fk(x) and gk(x) satisfy condition (3. 1) for ·each k.

Define functions

as follows:

vk(x) =

_f"0X'

^x

c -

_{1 -}^fk(x)) _x

wk(x)

=

_gk(x) ^x

^~l

_{1 -}^{- gk(x))} _x

Let

= _~

x

= _{gk(x) .} x

. . . ' r n ' . . .

be an enumeration of the rational numbers in (O, 1) and write

Now define sequences pk and qk as follows:

(k

=

1, 2, ••• ) • Since vk

2:

^uk

2:

1 and wk

2:

^zk

2:

^1, it follows that

and

.:: ~2k •

If for some k

>

1, both vk(x)

=

1 and wk(x) _ 1, then it fol- lows that

x

=

or

which implies fk(x)

=

x. Similarly, wk(x)

=

1 implies gk(x)

=

^x.

These relations in turn imply that R(x)

=

x and trR(x)

=

^{x since}

7k

x < x {k= 1,2, ••• ) .

Thus taking any covering classes S, T with S - T, would give R(S, T)

=

R .

Therefore, without loss of generality assume that for some k _;:: 1, there is a point x

0, 0 < x

0 < l, such that, vk(x

0);:: s > 1. There is then a

'neighborhood, U, of x s-1

0 on which vk(y) > 1

+ -

2- > 1. Since j _;:: k

implies

and so,

f.(x) < fk(x), it follows that J -

v.(y)

=

^{y (1-f/y))}

=

_Y_(l -f/y)) J

1jlY1

^{1 - y 1 - y f /Y)}

y (1 - fk(y)\ =

;:: 1 - y

l

^{f;k(y) )}

v .(y) > ¹ + - - > s-1 1 '

J - 2

for y in U and j _;:: k. Since r . is in U for infinitely many indices J.

J '

it follows that there are infinitely many j for which

and thus the product v 1 •••

fini ty. Since

v. approaches infj.nity as j approaches in- J

and since the sequence (pk) is non-increasing, it follows that pk tends to zero as k tends to infinity. Thus covering classes S and T can

be defined by writing:

s

= (pk)

u

(q2k) T = (pk) U (q2k- l)

(k=l,2, .•• ) (k=l,2, ••. ) The next step is to show that for 0 <a. < l

and

and this will complete the proof. Fir st consider the function (xa.)

8. At the points pk' (k = 1, 2, .•• )

since pk E

s.

^It remains to determine the V?-lues (xa.)

s<

q2k- l) (k

=

^1, 2, ••• ) • Sinc·e p 2k < q 2k- l .::_ p 2k- l' it follows that

by the remarkmade following Theorem l (§3,II). By definition,

and so that

vk 1-( 1-a) Uk q2k-l

Using the definitions of

Thus, for t ET,

log (xa.)

s<

^t)

log

t

Since

a. '

=

vk and uk and substituting, yields a.fk(rk) V (.1 - (l-a.)(1-fk(rk)

>)

rk 1-rk

=

^qZk-1

and

if tsp

k (k= 1, 2, .•• )

Since the rational numbers are everywhere dense in (0, 1), it follows that

which in turn implies

or

and thus

Similarly, if rk~a., then

( 1-a.)( 1-fk(r k) ) 1 - - - -1-r

k Hence

It is now easy to verify that

and

when 0 <a.< 1. Indeed, when rk >a.,

so that

is immediate.

On the other hand, given e

>

0, there is

o >

0, such that a. :::_ r k :::_ a.

+ o

implies

Since a:::_ rk:::_a.

+ o,

for arbitrarily large k and since fk(a.) tends to R(a.) as k tends to infinity, it follows that there is k, with

a:::_

^{rk and,}

and this shows that

= R(a.) •

Similarly, when rk

<a,

so again

> R(a.) •

An analogous argument shows that dS(

(xa.)~) =

^{trR(a.) when}

a<

1. Since R is closed, the assertion of the theorem follows by the Lemma above.

§3. A Characterization of R(S, T)

Let S and T be given covering classes. For each point t E T, there is associated a unique point (a(t), b(t)) in the unit square

12• This association is made as follows: Given· t, there are points

SO' sl in

s,

defined by

so = sup (s: s :::_ t, s ES) sl = ^inf ( s: s

2:.

t;', s E S)

where s 1 is defined to be 1, if there are no s E S with s

>

t. Write s0

= t°'

and s

1

=

tl3, where

a.

^{2:_ 12:_}

13·

Observe that if either

a.=

l or 13

=

l, then both are equal 1.

In this case define

a(t) = b(t) = 1/2, when

a.>

l

> 13,

the equations in a(t), b(t),

have the solution

a(t)

Altogether then, a(t) = l

2'

(3. 4)

b(t)

=

^{13 a(t) ,}

1 - b( t)

= a (

^1-a( t) )

= a-1

a-13 '

a. =

^{l ;}

b(t)

= 13(a-

l)

a-13

(t) ^{a-1 1}

a

=a.-13'

a>

b ( t)

= ~ • 13 . =

1 ; b ( t)

= 13 ~~-al)

^•

13

< 1 •

Note that a(t)

>

0 for all t E T and that b(t)

>

0 for t E T sufficiently small. Also, if

13 <

1 then both

and

13(a-

l) < ₁

a-13

- - < a-1 1 •

a.-13

Thus, in any event a(t)

<

l and b(t)

<

1, with a(t) ~ b(t), for all t E T.

It follows that

and

Define

= tb ( t)

I

a( t) sl

= t( 1-b(t)) /( 1-a(t))

so

An(S:T)

=

{ca(t). b(t) ):t E T. t:::.

~}

^•

and finally write,

(3. 5) A(S:T) =

n

A (S:T) ,

n=l n where A denotes the closure of A.

In a similar manner, there is .associated with each point s ES, a point (c(s), d(s)) in 1

2, such that

t ₀

=

1-c(s) l-d(s)

s '

c(s)

d\51

s

. .

where t 0, t

1 are the analogues of s 0, s

1 above. Then the set

(3. 6)

can be defined, where

A(T:S) = II n=l

A (T:S) n

An(T:S) = {(c(s), d{s) ):s E S, 's,:::

~}

^•

Since An(S:T)

?

An+l(S:T) and each of these sets is non- empty, it follows that the sets A(T:S) and A(S: T) are non-empty. The interest in the sets A(S:T) and A(T:S) is that while they are defined solely in terms of the points of S and T, the smallest doubly-starred

set containing A(S: T) U A( T:S) is precisely R(S, T), as the following results make clear.

Theorem 1. . (A(S: T)

U

A(T:S)

f =

^{R(S, T) •}

Proof: Since R(S, T) is doubly-starred, it is sufficient to show that

A(S: T) U A( T:S)

=

^{R(S, T) •}

The proof demonstrates only that

A(S: T) ::; R(S, T) ,

the proof that A( T:S)

:=;

R(S, T) being similar. Hence let

co

(a, b) E

n

A (S:T), and 0 < a< 1. Then there is a decreasing se- n=l n

quence (t ) c T, such that t -+ 0 and

n - n

a = a(t ) ... a

n n

(n ^{-+co) •} b = b(t ) ... b

n n

a 1-a Write s

0 ::::. tn::::. s

1 and put xn = s 1 s

0 , then for tn satisfying x < t <s

1, n - n -

and for s

0 < t < x ,

- n - n

(x~S(tn)

⁼

=

a-1 to

ab /a

=

^{t ' n n}

n

t n 1-b n

+ l (a-1)

1-a

t n

n

Thus in any case

and so

log t n

ab n

ab n

=

_{t n} ^{a n}

ab n a n

/\ t n

( 1-b ) 1

+

(a-1) (l-a) n

n

( 1-bn) V 1

+

(a-1) {1-a) n

1-b V 1

+

(a-1) _ _ n

1-a n

Since b(t ) ... b and a(t ) ... a as n ... co, it follows that

n n

which shows (a, b) E R(S, T). Now, if a= 0 .and (a, b) E A(S, T), then b = 0, and (0, 0) is clearly in R(S, T). If a= 1 and b = 1, then (a, b) is clearly in R(S, T). The only remaining case is a = 1, b < 1. Again let a(t ) ... a, b{t ) ... b t E T, t -+ O. Since b(t ) < 1 for n sufficiently

n n n n n

large, a(tn) < 1 for the same n. Let a.< 1 and consider (xa.)S(tn). As above

and

log {xa.)

s<

^tn)

---<

log t · n

a. b n

a n (

1-b ) V 1 - ( 1-a.) l-a:

Since a(t ) ... 1, it follows that n

log (xa.)

s<

^{tn) a b n}

- - - < - - .... ab,

log t a

n n

and the re fore dT( (xa.)

5) ~ab, for all a < 1. This implies

By the lemma of §2 it follows that (a, b) E R(S, T}.

Theorem 2. R(S, T) ~ (A(S: T)

U

A( T:S)) s

Proof: Let f E :J with d

5(f) ~ dT(f). If actually d

5(f)

=

dT(f) then (d

5(£), dT(f)) E (A(S: T)

U

A(T:S) )s automatically. Thus assume d

8(f)

>

dT(f). and let a, 13 be such that

By Lemma 1, (§1,I) for any t ET, s ES,

f(t)

~ (1

^V

!)

^{f(s) ,}

and so,

Since a< d

8(£), it follows by the Theorem of §1, Chapter II, that ( t E T) ( s E S) •

Since 13 > dT(f) there is a set T 0 ~ T, T 0 a covering class, such that t E T 0, implies

t-13f(t)

>

M • a.

Thus for s E S and t E T 0,

Choosing t E T

0, and s

0(t) _::: t_::: s

1(t). (si(t) ES) (i = 0, 1) , it follows that

and

Sl(t) -- tb{t)/a(t) and so(t) -- tl-b(t)/1-a(t). h 1

Since t ese equa ities imply

that

-13 +a. b( t) l < t affj and

1-13

+

(a.-1) (1-b(t)j

1 < t 1-a(ijj .

Since t _::: 1, it follows that b(t) < ].

anr -

^{a. and}

^{T=a _:::}

^{1-13 1-b( t)}

^1-

^{a( t)}

Let (a, b) E A(S, T

0) ~ A(S, _T) and let tn be a 'sequence in T

0 such that

Then it follows that

a(t ) ... a, b(t ) ... b •

n n

~ < ].

a

a.

^and

1-13 1-b

<

1-a. - 1-a

s s s which implies that (a., 13) E A(S, T

0) ~ A(S:T) ::; (A(S:T) U A(T:S)) . Since (a., 13) can be chosen arbitrarily close to (d

5(£), dT(f)) it follows that (d

8(f), dT(f)) E (A(S: T) U A(T:S)) s, this latter set being closed by Lemma 1, § 1. The case d

8(£) < dT(f) is treated similarly, and so the proof is complete.

Theorems l and 2 combine to give

Theorem 3. R(S, T)

=

(A(S: T) U A(T:S)) s •