Where do

informationalpartitions come from?

Tho saugh I ... alle the mervelous signalsOf the goddys celestials Chaucer, The House of Fame (c1384)

.

....

..

StatesStates

Signal = player’s information = player’s knowledge

Signal = player’s information = player’s knowledge

.

....

..

E

From partition to knowledge

K(E)

Here the player knows the element of the partition that contains the states.

...and also E.

The event that she knows E.

Partitional knowledge

Partitional knowledge

ω

(ω)

Ω – a state space – a partition of Ω(ω) – the element of that contains state ω. ω K(E) when (ω) E.

Ω – a state space – a partition of Ω(ω) – the element of that contains state ω. ω K(E) when (ω) E.

From knowledge to partitionSyntax

Bob knows that G.W. Bush is the president of the US and he does not know that Barbara is G.W.’s wife.

The set of all sentences form a Boolean algebra (w.r.t “and”, “or”, “not”) with an operator (know)

Language:

An algebra of subsets, Awith an operator K: A A.

Language:

An algebra of subsets, Awith an operator K: A A.

A family of sets closed under intersection and complement

Sentences

Deduction rules

vs. Semantics

From knowledge to partition

The axioms of S5 knowledge:

0. The axioms of Boolean algebra

1. K() =

2. K(E) K(F) = K(E F)

3. K(E) E

4. ¬ K(E) = K(¬ K(E))

0. The axioms of Boolean algebra

1. K() =

2. K(E) K(F) = K(E F)

3. K(E) E

4. ¬ K(E) = K(¬ K(E))

British Panel Gives Rumsfeld 'Foot In Mouth' Award

Defense Secretary's Comment On 'Known Unknowns' Is Most Baffling

LONDON -- He may not know it -- or he may know that he knows it -- but Secretary of Defense Donald Rumsfeld has won this year's "Foot in Mouth" award for the by a public figure.

Britain's Plain English Campaign cited Rumsfeld's comment on Iraq, when he said the following during a Pentagon briefing:

"Reports that say that something hasn't happened are always interesting to me, because as we know, , there are things we know we know.

"We also know . That is to say we know there are some things we do not know. But there are also

most baffling statement

there are known knowns

there are known unknowns unknown unknowns: the oneswe don't know we don't know.”

From knowledge to partition

The axioms of S5 knowledge:

0. The axioms of Boolean algebra

1. K() =

2. K(E) K(F) = K(E F)

3. K(E) E

4. ¬ K(E) = K(¬ K(E))

0. The axioms of Boolean algebra

1. K() =

2. K(E) K(F) = K(E F)

3. K(E) E

4. ¬ K(E) = K(¬ K(E))

Is S5-knowledge partitional?Partitional knowledge satisfies S5.

Fairy tale:

S5 implies partition

Fairy tale:

S5 implies partition

Hard facts:

Words alone do not bring about partitions

Hard facts:

Words alone do not bring about partitions

.S5

... . .. .-3 -2 -1 0 1 2 3

. . . . . . . ...4 5-5 -4

A0 – the algebra generated by all arithmetic sequences.

An arithmetic sequence: {a + zd | z Z} for d 0.

Ω = the set of integers, Z.

Example 1

e.g. -2 + 3z

K: A0 A0 is the identity.

K is generated by the partition into singletons.

... . .. .-3 -2 -1 0 1 2 3

. . . . . . . ...4 5-5 -4

A – the algebra generated by A0 and the set P.

K(E) = F

P

Example 1

Each E A can be uniquely decomposed:

E = X F Y

P ¬ P

A0

1. K() =

2. K(E) K(E’) = K(E E’)

3. K(E) E

4. ¬ K(E) = K(¬ K(E))

K is the identity on A0

K (P) =

2K satisfies S5.

The candidate for a partition: singletons.

If K is partitional then at state 3 the player knows P (because it contains {3}).

Therefore K is not partitional.

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From partition to knowledges common

1 -

2 -

c - coarser than 1 and 2

finest than any such partition

Kc - narower than K1 and K2

For each E, Kc(E) K1(E), K2(E)

broader than any such knowledge

the common knowledge partition.

Agreeing to disagree, Aumann (1976)

From partition to knowledges common

1 -

2 -

c - coarser than 1 and 2

finest than any such partition

Kc - narower than K1 and K2

For each E, Kc(E) K1(E), K2(E)

broader than any such knowledge

the common knowledge partition.

Agreeing to disagree, Aumann (1976)

Is it possible to prove syntactically the existence of Kc for S5-knowledge K1 and K2 ?

Is it possible to prove syntactically the existence of Kc for S5-knowledge K1 and K2 ?

NO!

Is there a set S of axioms of knowledge such that

K satisfies S K is partitional

For S = S5:

For S = {K(E) = E}: NO!