How to show $operatorname{card}(omega+1)=omega$
$begingroup$
Apparently $operatorname{card}(omega+1)=omega$. This means that there is an order $<$ on $omega+1$ such that there is an isomorphism of ordered set $f$, $(omega+1,<) cong (omega,in)$, but to what does $f$ send $omega in omega+1$?
Edit
Also to show $card(omega+1)=omega$ I think I need to show that any ordinal $alpha$ that is in bijection with $omega+1$ is such that $alpha ge omega$ no? [I know also that if two ordinals are isomorphic ($langle beta, in rangle cong langle gamma, in rangle$ i.e. isomorphism of ordered set) then $beta=gamma$, is this usefull here?.]
elementary-set-theory cardinals
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
$endgroup$
add a comment |
$begingroup$
Apparently $operatorname{card}(omega+1)=omega$. This means that there is an order $<$ on $omega+1$ such that there is an isomorphism of ordered set $f$, $(omega+1,<) cong (omega,in)$, but to what does $f$ send $omega in omega+1$?
Edit
Also to show $card(omega+1)=omega$ I think I need to show that any ordinal $alpha$ that is in bijection with $omega+1$ is such that $alpha ge omega$ no? [I know also that if two ordinals are isomorphic ($langle beta, in rangle cong langle gamma, in rangle$ i.e. isomorphism of ordered set) then $beta=gamma$, is this usefull here?.]
elementary-set-theory cardinals
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
$endgroup$
add a comment |
$begingroup$
Apparently $operatorname{card}(omega+1)=omega$. This means that there is an order $<$ on $omega+1$ such that there is an isomorphism of ordered set $f$, $(omega+1,<) cong (omega,in)$, but to what does $f$ send $omega in omega+1$?
Edit
Also to show $card(omega+1)=omega$ I think I need to show that any ordinal $alpha$ that is in bijection with $omega+1$ is such that $alpha ge omega$ no? [I know also that if two ordinals are isomorphic ($langle beta, in rangle cong langle gamma, in rangle$ i.e. isomorphism of ordered set) then $beta=gamma$, is this usefull here?.]
elementary-set-theory cardinals
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
$endgroup$
Apparently $operatorname{card}(omega+1)=omega$. This means that there is an order $<$ on $omega+1$ such that there is an isomorphism of ordered set $f$, $(omega+1,<) cong (omega,in)$, but to what does $f$ send $omega in omega+1$?
Edit
Also to show $card(omega+1)=omega$ I think I need to show that any ordinal $alpha$ that is in bijection with $omega+1$ is such that $alpha ge omega$ no? [I know also that if two ordinals are isomorphic ($langle beta, in rangle cong langle gamma, in rangle$ i.e. isomorphism of ordered set) then $beta=gamma$, is this usefull here?.]
elementary-set-theory cardinals
elementary-set-theory cardinals
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
edited Jan 22 at 15:09
roi_saumon
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
asked Jan 22 at 14:29
roi_saumonroi_saumon
56738
56738
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
If it it only concerns cardinality then a bijection is enough.
If you insist on an order preserving bijection then:
Let $<$ on $omega+1$ be defined by $omega<0$ and $n<m$ if $n,minomega$ with $nin m$.
Then $f$ can be prescribed by:
- $omegamapsto0$
$nmapsto n+1$ for $ninomega$
answered Jan 22 at 14:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
add a comment |
$begingroup$
This is know as the Hilbert hotel paradox.
The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $omega$.
$1$ new customer arrives, but the hotel is full, every room is occupied by $omega$ customers.
But the butler just invites the new customer to take the first room (labelled $1$) and asks the initial occupant to move to the next room ($1mapsto 2$) and so on ($nmapsto n+1$) for successive occupants.
Finally $omega+1$ customers can fit in the Hilbert hotel.
answered Jan 22 at 15:24
zwimzwim
12.5k831
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If it it only concerns cardinality then a bijection is enough.
If you insist on an order preserving bijection then:
Let $<$ on $omega+1$ be defined by $omega<0$ and $n<m$ if $n,minomega$ with $nin m$.
Then $f$ can be prescribed by:
- $omegamapsto0$
$nmapsto n+1$ for $ninomega$
answered Jan 22 at 14:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
add a comment |
$begingroup$
If it it only concerns cardinality then a bijection is enough.
If you insist on an order preserving bijection then:
Let $<$ on $omega+1$ be defined by $omega<0$ and $n<m$ if $n,minomega$ with $nin m$.
Then $f$ can be prescribed by:
- $omegamapsto0$
$nmapsto n+1$ for $ninomega$
answered Jan 22 at 14:37
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
add a comment |
$begingroup$
If it it only concerns cardinality then a bijection is enough.
If you insist on an order preserving bijection then:
Let $<$ on $omega+1$ be defined by $omega<0$ and $n<m$ if $n,minomega$ with $nin m$.
Then $f$ can be prescribed by:
- $omegamapsto0$
$nmapsto n+1$ for $ninomega$
answered Jan 22 at 14:37
drhabdrhab
102k545136
$endgroup$
If it it only concerns cardinality then a bijection is enough.
If you insist on an order preserving bijection then:
Let $<$ on $omega+1$ be defined by $omega<0$ and $n<m$ if $n,minomega$ with $nin m$.
Then $f$ can be prescribed by:
- $omegamapsto0$
$nmapsto n+1$ for $ninomega$
answered Jan 22 at 14:37
drhabdrhab
102k545136
edited Jan 22 at 14:41
answered Jan 22 at 14:37
drhabdrhab
102k545136
answered Jan 22 at 14:37
drhabdrhab
102k545136
answered Jan 22 at 14:37
drhabdrhab
102k545136
102k545136
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
add a comment |
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
$begingroup$
Thanks, I forgot one thing though that I edited in the question.
$endgroup$
– roi_saumon
Jan 22 at 15:10
add a comment |
$begingroup$
This is know as the Hilbert hotel paradox.
The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $omega$.
$1$ new customer arrives, but the hotel is full, every room is occupied by $omega$ customers.
But the butler just invites the new customer to take the first room (labelled $1$) and asks the initial occupant to move to the next room ($1mapsto 2$) and so on ($nmapsto n+1$) for successive occupants.
Finally $omega+1$ customers can fit in the Hilbert hotel.
answered Jan 22 at 15:24
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
This is know as the Hilbert hotel paradox.
The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $omega$.
$1$ new customer arrives, but the hotel is full, every room is occupied by $omega$ customers.
But the butler just invites the new customer to take the first room (labelled $1$) and asks the initial occupant to move to the next room ($1mapsto 2$) and so on ($nmapsto n+1$) for successive occupants.
Finally $omega+1$ customers can fit in the Hilbert hotel.
answered Jan 22 at 15:24
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
This is know as the Hilbert hotel paradox.
The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $omega$.
$1$ new customer arrives, but the hotel is full, every room is occupied by $omega$ customers.
But the butler just invites the new customer to take the first room (labelled $1$) and asks the initial occupant to move to the next room ($1mapsto 2$) and so on ($nmapsto n+1$) for successive occupants.
Finally $omega+1$ customers can fit in the Hilbert hotel.
answered Jan 22 at 15:24
zwimzwim
12.5k831
$endgroup$
This is know as the Hilbert hotel paradox.
The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $omega$.
$1$ new customer arrives, but the hotel is full, every room is occupied by $omega$ customers.
But the butler just invites the new customer to take the first room (labelled $1$) and asks the initial occupant to move to the next room ($1mapsto 2$) and so on ($nmapsto n+1$) for successive occupants.
Finally $omega+1$ customers can fit in the Hilbert hotel.
answered Jan 22 at 15:24
zwimzwim
12.5k831
answered Jan 22 at 15:24
zwimzwim
12.5k831
answered Jan 22 at 15:24
zwimzwim
12.5k831
answered Jan 22 at 15:24
zwimzwim
12.5k831
12.5k831
add a comment |
add a comment |
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