Metamath Proof Explorer


Theorem alephom

Description: From canth2 , we know that ( aleph0 ) < ( 2 ^om ) , but we cannot prove that ( 2 ^ om ) = ( aleph1 ) (this is the Continuum Hypothesis), nor can we prove that it is less than any bound whatsoever (i.e. the statement ( alephA ) < ( 2 ^om ) is consistent for any ordinal A ). However, we can prove that ( 2 ^ om ) is not equal to ( aleph_om ) , nor ( aleph( aleph_om ) ) , on cofinality grounds, because by Konig's Theorem konigth (in the form of cfpwsdom ), ( 2 ^om ) has uncountable cofinality, which eliminates limit alephs like ( alephom ) . (The first limit aleph that is not eliminated is ( aleph( aleph1 ) ) , which has cofinality ( aleph1 ) .) (Contributed by Mario Carneiro, 21-Mar-2013)

Ref Expression
Assertion alephom card 2 𝑜 ω ω

Proof

Step Hyp Ref Expression
1 sdomirr ¬ ω ω
2 2onn 2 𝑜 ω
3 2 elexi 2 𝑜 V
4 domrefg 2 𝑜 V 2 𝑜 2 𝑜
5 3 cfpwsdom 2 𝑜 2 𝑜 cf card 2 𝑜
6 3 4 5 mp2b cf card 2 𝑜
7 aleph0 = ω
8 7 a1i card 2 𝑜 ω = ω = ω
9 7 oveq2i 2 𝑜 = 2 𝑜 ω
10 9 fveq2i card 2 𝑜 = card 2 𝑜 ω
11 10 eqeq1i card 2 𝑜 = ω card 2 𝑜 ω = ω
12 11 biimpri card 2 𝑜 ω = ω card 2 𝑜 = ω
13 12 fveq2d card 2 𝑜 ω = ω cf card 2 𝑜 = cf ω
14 limom Lim ω
15 alephsing Lim ω cf ω = cf ω
16 14 15 ax-mp cf ω = cf ω
17 cfom cf ω = ω
18 16 17 eqtri cf ω = ω
19 13 18 eqtrdi card 2 𝑜 ω = ω cf card 2 𝑜 = ω
20 8 19 breq12d card 2 𝑜 ω = ω cf card 2 𝑜 ω ω
21 6 20 mpbii card 2 𝑜 ω = ω ω ω
22 21 necon3bi ¬ ω ω card 2 𝑜 ω ω
23 1 22 ax-mp card 2 𝑜 ω ω