If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal
$begingroup$
If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.
My attempt:
Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.
It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.
My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!
proof-verification elementary-set-theory ordinals
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
add a comment |
$begingroup$
If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.
My attempt:
Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.
It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.
My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!
proof-verification elementary-set-theory ordinals
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
add a comment |
$begingroup$
If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.
My attempt:
Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.
It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.
My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!
proof-verification elementary-set-theory ordinals
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.
My attempt:
Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.
It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.
My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!
proof-verification elementary-set-theory ordinals
proof-verification elementary-set-theory ordinals
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
asked Dec 9 '18 at 1:24
Le Anh DungLe Anh Dung
1,4141621
1,4141621
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes your argument follows. In general, any infinite cardinal is a limit ordinal.
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
$endgroup$
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes your argument follows. In general, any infinite cardinal is a limit ordinal.
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
$endgroup$
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
add a comment |
$begingroup$
Yes your argument follows. In general, any infinite cardinal is a limit ordinal.
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
$endgroup$
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
add a comment |
$begingroup$
Yes your argument follows. In general, any infinite cardinal is a limit ordinal.
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
$endgroup$
Yes your argument follows. In general, any infinite cardinal is a limit ordinal.
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
answered Dec 9 '18 at 1:28
Alberto TakaseAlberto Takase
2,380719
2,380719
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
add a comment |
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
$begingroup$
Oh i forget that. Many thanks!
$endgroup$
– Le Anh Dung
Dec 9 '18 at 1:29
add a comment |
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