idempotency of infinite cardinals

idempotency of infinite cardinals

In this entry, we show that every

infinite

cardinal

is

idempotent

with respect to

cardinal addition

and

cardinal multiplication

.

Theorem 1

.

κ

⋅

κ

=

κ

for any infinite cardinal

κ

.

Proof.

For any non-zero cardinal

λ

, we have

λ

=

1

⋅

λ

≤

λ

⋅

λ

. So given an infinite cardinal

κ

, either

κ

=

κ

⋅

κ

or

κ

<

κ

⋅

κ

. Let

𝒞

be the class of infinite cardinals that fail to be idempotent (with respect to

⋅

). Suppose

𝒞

≠

∅

. We shall derive a

contradiction

. Since

𝒞

consists entirely of

ordinals

, it is therefore

well-ordered

, and has a least member

κ

.

Let

K

=

κ

×

κ

. As

K

is a

collection

of

ordered pairs

of ordinals, it has the

canonical well-ordering

inherited from the canonical ordering on

On

×

On

. Let

α

be the ordinal

isomorphic

to

K

. Since

κ

<

κ

⋅

κ

=

|

K

|

, there is an

initial segment

L

of

K

that is

order isomorphic

to

κ

.

Since

L

is an initial segment of

K

,

L

=

{

(

β

1

,

β

2

)

∣

(

β

1

,

β

2

)

≺

(

α

1

,

α

2

)

}

for some

(

α

1

,

α

2

)

∈

K

. The well-order

⪯

denotes the canonical ordering on

K

. Let

λ

=

max

⁡

(

α

1

,

α

2

)

. Since

L

⊂

K

=

κ

×

κ

,

α

1

<

κ

and

α

2

<

κ

, and therefore

λ

<

κ

.

For any

(

β

1

,

β

2

)

∈

L

, we have

(

β

1

,

β

2

)

≺

(

α

1

,

α

2

)

, which implies that

max

⁡

(

β

1

,

β

2

)

≤

λ

. Therefore

L

⊆

λ

+

×

λ

+

, or

|

L

|

≤

|

λ

+

×

λ

+

|

≤

|

λ

+

|

⋅

|

λ

+

|

. There are two cases to discuss:

1.

If

λ

is finite, so is

λ

+

×

λ

+

, contradicting that

L

is (order) isomorphic to

κ

, an

infinite set

.

2.

If

λ

is infinite, so is

|

λ

+

|

. Since

λ

<

κ

, and

κ

is a

limit ordinal

,

|

λ

+

|

<

k

as well, which means

|

λ

+

|

∉

𝒞

, or

|

λ

+

|

⋅

|

λ

+

|

=

|

λ

+

|

. Therefore

|

L

|

≤

|

λ

+

|

⋅

|

λ

+

|

=

|

λ

+

|

≤

λ

+

<

κ

, again contradicting that

L

is (order) isomorphic to

κ

.

Therefore, the

assumption

𝒞

≠

∅

is false, and the proof is

complete

.
∎

Corollary 1

.

If

0

<

λ

≤

κ

and

κ

is infinite, then

λ

⋅

κ

=

κ

.

Proof.

κ

=

1

⋅

κ

≤

λ

⋅

κ

≤

κ

⋅

κ

=

κ

. By Schroder-Bernstein’s

Theorem

,

λ

⋅

κ

=

κ

.
∎

Corollary 2

.

If

λ

≤

κ

and

κ

is infinite, then

λ

+

κ

=

κ

.

Proof.

κ

=

0

+

κ

≤

λ

+

κ

≤

κ

+

κ

=

2

⋅

κ

≤

κ

⋅

κ

=

κ

by the corollary above (since

2

≤

κ

). Another application of Schroder-Bernstein gives

κ

=

λ

+

κ

.
∎

Since

κ

≤

κ

, we get the following:

Corollary 3

.

κ

+

κ

=

κ

for any infinite cardinal.

Remark

. No cardinal greater than

1

is idempotent with respect to

cardinal exponentiation

. This is a direct

consequence

of Cantor’s theorem:

κ

<

2

κ

≤

κ

κ

.

Title

idempotency of infinite cardinals

Canonical name

IdempotencyOfInfiniteCardinals

Date of creation

2013-03-22 18:53:30

Last modified on

2013-03-22 18:53:30

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03E10

Related topic

CanonicalWellOrdering

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