Mixing
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
59438
$endgroup$
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
59438
$endgroup$
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
59438
$endgroup$
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
59438
asked Jan 22 at 14:29
roi_saumonroi_saumon
59438
edited Jan 22 at 15:09
roi_saumon
asked Jan 22 at 14:29
roi_saumonroi_saumon
59438
asked Jan 22 at 14:29
roi_saumonroi_saumon
59438
asked Jan 22 at 14:29
roi_saumonroi_saumon
59438
59438
add a comment |
add a comment |
2 Answers
2
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
103k545136
$endgroup$
$begingroup$
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
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083233%2fhow-to-show-operatornamecard-omega1-omega%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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
103k545136
$endgroup$
$begingroup$
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
103k545136
$endgroup$
$begingroup$
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
103k545136
$endgroup$
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
103k545136
edited Jan 22 at 14:41
answered Jan 22 at 14:37
drhabdrhab
103k545136
answered Jan 22 at 14:37
drhabdrhab
103k545136
answered Jan 22 at 14:37
drhabdrhab
103k545136
103k545136
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083233%2fhow-to-show-operatornamecard-omega1-omega%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
