proof of ◇ is equivalent to ♣ and continuum hypothesis

proof of

◇

is equivalent to

♣

and continuum hypothesis

The proof that

◇

S

implies both

♣

S

and that for every

λ

<

κ

,

2

λ

≤

κ

are given in the entries for

◇

S

and

♣

S

.

Let

A

=

⟨

A

α

⟩

α

∈

S

be a

sequence

which

satisfies

♣

S

.

Since there are only

κ

bounded

subsets of

κ

, there is a surjective function

f

:

κ

→

Bounded

⁡

(

κ

)

×

κ

where

Bounded

⁡

(

κ

)

is the bounded subsets of

κ

. Define a sequence

B

=

⟨

B

α

⟩

α

<

κ

by

B

α

=

f

⁢

(

α

)

if

s

⁢

u

⁢

p

⁢

(

B

α

)

<

α

and

∅

otherwise. Since the set of

(

B

α

,

λ

)

∈

Bounded

⁡

(

κ

)

×

κ

such that

B

α

=

T

is

unbounded

for any bounded subset

T

, it follow that every bounded subset of

κ

occurs

κ

times in

B

.

We can define a new sequence,

D

=

⟨

D

α

⟩

α

∈

S

such that

x

∈

D

α

↔

x

∈

B

β

for some

β

∈

A

α

. We can show that

D

satisfies

◇

S

.

First, for any

α

,

x

∈

D

α

means that

x

∈

B

β

for some

β

∈

A

α

, and since

B

β

⊆

β

∈

A

α

⊆

α

, we have

D

α

⊆

α

.

Next take any

T

⊆

κ

. We consider two cases:

T

is bounded

The set of

α

such that

T

=

B

α

forms an unbounded sequence

T

′

, so there is a

stationary

S

′

⊆

S

such that

α

∈

S

′

↔

A

α

⊂

T

′

. For each such

α

,

x

∈

D

α

↔

x

∈

B

i

for some

i

∈

A

α

⊂

T

′

. But each such

B

i

is equal to

T

, so

D

α

=

T

.

T

is unbounded

We define a

function

j

:

κ

→

κ

as follows:

•

j

⁢

(

0

)

=

0

•

To find

j

⁢

(

α

)

, take

X

∩

{

j

⁢

(

β

)

∣

β

<

α

}

. This is a bounded subset of

κ

, so is equal to an unbounded series of elements of

B

. Take

j

⁢

(

α

)

=

γ

, where

γ

is the least number greater than any element of

{

α

}

∪

{

j

⁢

(

β

)

∣

β

<

α

}

such that

B

γ

=

X

∩

{

j

⁢

(

β

)

∣

β

<

α

}

.

Let

T

′

=

range

⁡

(

j

)

. This is obviously unbounded, and so there is a stationary

S

′

⊆

S

such that

α

∈

S

′

↔

A

α

⊆

T

′

.

Next, consider

C

, the set of

ordinals

less than

κ

closed under

j

. Clearly it is unbounded, since if

λ

<

κ

then

j

⁢

(

λ

)

includes

j

⁢

(

α

)

for

α

<

λ

, and so

induction

gives an ordinal greater than

λ

closed under

j

(essentially the result of applying

j

an

infinite

number of times). Also,

C

is closed: take any

c

⊆

C

and suppose

sup

⁡

(

c

∩

α

)

=

α

. Then for any

β

<

α

, there is some

γ

∈

c

such that

β

<

γ

<

α

and therefore

j

⁢

(

β

)

<

γ

. So

α

is closed under

j

, and therefore

contained

in

C

.

Since

C

is a club,

C

′

=

C

∩

S

′

is stationary. Suppose

α

∈

C

′

. Then

x

∈

D

α

↔

x

∈

B

β

where

β

∈

A

α

. Since

α

∈

S

′

,

β

∈

range

⁡

(

j

)

, and therefore

B

β

⊆

T

. Next take any

x

∈

T

∩

α

. Since

α

∈

C

, it is closed under

j

, hence there is some

γ

∈

α

such that

j

⁢

(

x

)

∈

γ

. Since

sup

⁡

(

A

α

)

=

α

, there is some

η

∈

A

α

such that

γ

<

η

, so

j

⁢

(

x

)

∈

η

. Since

η

∈

A

α

,

B

η

⊆

D

α

, and since

η

∈

range

⁡

(

j

)

,

j

⁢

(

δ

)

∈

B

η

for any

δ

<

j

-

1

⁢

(

η

)

, and in particular

x

∈

B

η

. Since we showed above that

D

α

⊆

α

, we have

D

α

=

T

∩

α

for any

α

∈

C

′

.

Title

proof of

◇

is

equivalent

to

♣

and

continuum hypothesis

Canonical name

ProofOfDiamondIsEquivalentToclubsuitAndContinuumHypothesis

Date of creation

2013-03-22 12:53:57

Last modified on

2013-03-22 12:53:57

Owner

Henry (455)

Last modified by

Henry (455)

Numerical id

6

Author

Henry (455)

Entry type

Proof

Classification

msc 03E45

Synonym

proof that diamond is equivalent to club and continuum hypothesis

Related topic

Diamond

Related topic

Clubsuit

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