proof of alternative characterization of ultrafilter

proof of alternative characterization of ultrafilter

Proof that

A

⁢

∐

B

=

X

implies

A

∈

𝒰

or

B

∈

𝒰

Once we show that

A

∉

𝒰

implies

B

∉

𝒰

, this result will follow immediately.

On the one hand, suppose that

A

∉

𝒰

and that there exists a

C

∈

𝒰

such that

A

∩

C

is empty. Then

C

⊆

B

. Since

𝒰

is a filter and

C

∈

𝒰

, this implies that

B

∈

𝒰

.

On the other hand, suppose that

A

∉

𝒰

and that

A

∩

C

is not empty for any

C

in

𝒰

. Then

{

A

}

∪

𝒰

would be a

filter subbasis

. The filter which it would generate would be finer than

𝒰

. The fact that

𝒰

is an

ultrafilter

means that there exists no filter finer than

𝒰

. This

contradiction

shows that, if

A

∉

𝒰

, then there exists a

C

such that

A

∩

C

is empty. But this would imply that

C

⊆

B

which, in turn would imply that

B

∈

𝒰

.

Proof that

𝒰

is an ultrafilter.

Assume that

𝒰

is a filter, but not an ultrafilter and that

A

⁢

∐

B

=

X

implies

A

∈

𝒰

or

B

∈

𝒰

. Since

𝒰

is not an ultrafilter, there must exist filter

𝒰

′

which is

strictly finer

. Hence there must exist

A

∈

𝒰

′

such that

A

∉

𝒰

. Set

B

=

X

∖

A

. Since

A

⁢

∐

B

=

X

and

A

∉

𝒰

, it follows that

B

∈

𝒰

. Since

𝒰

⊂

𝒰

′

, it is also the case that

B

∈

𝒰

′

. But

A

∈

𝒰

′

as well; since

𝒰

′

is a filter,

A

∩

B

∈

𝒰

′

. This is impossible because

A

∩

B

∈

𝒰

′

is empty. Therefore, no such filter

𝒰

′

can exist and

𝒰

must be an ultrafilter.

Proof of generalization to

A

∪

B

=

X

On the one hand, since

A

∪

B

=

X

implies

A

⁢

∐

B

=

X

, the condition

A

∪

B

=

X

⇒

A

∈

𝒰

∨

B

∈

𝒰

will also imply that

𝒰

is an ultrafilter.

On the other hand, if

A

∪

B

=

X

, there must exists

A

′

⊆

A

and

B

′

⊆

B

such that

A

′

⁢

∐

B

′

=

X

. If

𝒰

is assumed to be a filter,

A

′

∈

𝒰

implies that

A

∈

𝒰

. Likewise,

B

′

∈

𝒰

implies that

B

∈

𝒰

. Hence, if

𝒰

is a filter such that

A

∪

B

=

X

implies that either

A

∈

𝒰

or

B

∈

𝒰

, then

𝒰

is an ultrafilter.

Proof of first proposition regarding finite unions

Let

B

j

=

∐

i

=

1

j

A

i

and let

C

j

=

∐

i

=

j

+

1

n

A

i

. For each

i

between

1

and

n

-

1

, we have

B

i

⁢

∐

C

i

=

X

. Hence, either

B

i

∈

𝒰

or

C

i

∈

𝒰

for each

i

between

1

and

n

-

1

. Next, consider three possibilities:

1.

B

1

∈

𝒰

: Since

B

1

=

A

1

, it follows that

A

1

∈

𝒰

.

2.

B

n

-

1

∉

𝒰

: Since

B

n

-

1

⁢

∐

C

n

-

1

=

X

, it follows that

C

n

-

1

∈

𝒰

. Because

C

n

-

1

=

A

n

, it follows that

A

n

∈

𝒰

.

3.

B

1

∉

𝒰

and

B

n

-

1

∈

𝒰

: There must exist an

i

∈

{

2

,

…

,

n

-

1

}

such that

B

i

-

1

∉

𝒰

and

B

i

∈

𝒰

. Since

B

i

-

1

∉

𝒰

,

C

i

-

1

∈

𝒰

.
Since

𝒰

is a filter,

C

i

-

1

∩

B

i

∈

𝒰

. But also

C

i

-

1

∩

B

i

=

A

i

which implies that

A

i

∈

𝒰

.

This examination of cases shows that if

∐

i

=

1

n

A

i

=

X

, then there must exist an

i

such that

A

i

∈

𝒰

. It is also easy to see that this

i

is unique — If

A

i

∈

𝒰

and

A

j

∈

𝒰

and

i

≠

j

, then

A

i

∩

A

j

=

∅

, but this cannot be the case since

𝒰

is a filter.

Proof of second proposition regarding finite unions

There exist sets

A

i

′

such that

A

i

′

⊆

A

i

and

∐

i

=

1

n

A

i

=

X

. By the result just proven, there exists an

i

such that

A

i

′

∈

𝒰

. Since

𝒰

is a filter,

A

i

′

∈

𝒰

implies

A

i

∈

𝒰

. Note that we can no longer assert that

i

is unique because the

A

i

’s no longer are required to be

pairwise disjoint

.

Title

proof of alternative characterization of ultrafilter

Canonical name

ProofOfAlternativeCharacterizationOfUltrafilter

Date of creation

2013-03-22 14:42:23

Last modified on

2013-03-22 14:42:23

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

16

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 54A20

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