$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$) Announcing the...
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$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Low Level Well-Ordering Principle ProofProving that well ordering principle implies Zorn's Lemma.well ordering principle implies Zorn's LemmaFind a set A of rational numbers that is isomorphic to sup{$omega$, $omega^omega$, $omega^{omega^omega}$, $cdots$}Proving the well ordering principleProving with well ordering principleFormal statement of well-ordering (not the theorem)Exercise: prove the well-ordering principle using Zorn's lemmaDoes the proof of the well-ordering principle not commit the circular reasoning fallacy?Why is partial ordering by continuation (vs just inclusion) required for Zorn's Lemma $implies$ Well Ordering Theorem?
$begingroup$
$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$).
I see that $omega +1$ does have maximal element but $omega$ is not so there is no ismorphism between $omega +1 $ and $omega $ but how can I write as proof this?
elementary-set-theory proof-writing
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
$endgroup$
add a comment |
$begingroup$
$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$).
I see that $omega +1$ does have maximal element but $omega$ is not so there is no ismorphism between $omega +1 $ and $omega $ but how can I write as proof this?
elementary-set-theory proof-writing
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
$endgroup$
2
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
2
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09
add a comment |
$begingroup$
$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$).
I see that $omega +1$ does have maximal element but $omega$ is not so there is no ismorphism between $omega +1 $ and $omega $ but how can I write as proof this?
elementary-set-theory proof-writing
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
$endgroup$
$omega +1 $ is not isomorphic to $omega$ (in the well-ordering by $varepsilon$).
I see that $omega +1$ does have maximal element but $omega$ is not so there is no ismorphism between $omega +1 $ and $omega $ but how can I write as proof this?
elementary-set-theory proof-writing
elementary-set-theory proof-writing
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
asked Mar 23 at 22:40
PozcuKushimotoStreetPozcuKushimotoStreet
1,409923
1,409923
2
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
2
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09
add a comment |
2
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
2
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09
2
2
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
2
2
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following is how you might translate your reasoning into a slightly more formal proof.
Suppose towards contradiction that $varphicolon omega+1to omega$ is an isomorphism. Notice that $omegain omega+1$ is a maximal element.
Since order isomorphisms map maximal elements to maximal elements, it follows that $varphi(omega)$ is maximal in $omega$. But $omega$ has no maximal element, a contradiction, which finishes the proof.
As suggested by Henno Brandsma in the comments, this proof presupposes that we already know that being a maximal element is preserved by order isomorphism. If that is not completely clear, the proof of this fact should be included.
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
$endgroup$
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
add a comment |
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$begingroup$
The following is how you might translate your reasoning into a slightly more formal proof.
Suppose towards contradiction that $varphicolon omega+1to omega$ is an isomorphism. Notice that $omegain omega+1$ is a maximal element.
Since order isomorphisms map maximal elements to maximal elements, it follows that $varphi(omega)$ is maximal in $omega$. But $omega$ has no maximal element, a contradiction, which finishes the proof.
As suggested by Henno Brandsma in the comments, this proof presupposes that we already know that being a maximal element is preserved by order isomorphism. If that is not completely clear, the proof of this fact should be included.
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
$endgroup$
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
add a comment |
$begingroup$
The following is how you might translate your reasoning into a slightly more formal proof.
Suppose towards contradiction that $varphicolon omega+1to omega$ is an isomorphism. Notice that $omegain omega+1$ is a maximal element.
Since order isomorphisms map maximal elements to maximal elements, it follows that $varphi(omega)$ is maximal in $omega$. But $omega$ has no maximal element, a contradiction, which finishes the proof.
As suggested by Henno Brandsma in the comments, this proof presupposes that we already know that being a maximal element is preserved by order isomorphism. If that is not completely clear, the proof of this fact should be included.
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
$endgroup$
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
add a comment |
$begingroup$
The following is how you might translate your reasoning into a slightly more formal proof.
Suppose towards contradiction that $varphicolon omega+1to omega$ is an isomorphism. Notice that $omegain omega+1$ is a maximal element.
Since order isomorphisms map maximal elements to maximal elements, it follows that $varphi(omega)$ is maximal in $omega$. But $omega$ has no maximal element, a contradiction, which finishes the proof.
As suggested by Henno Brandsma in the comments, this proof presupposes that we already know that being a maximal element is preserved by order isomorphism. If that is not completely clear, the proof of this fact should be included.
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
$endgroup$
The following is how you might translate your reasoning into a slightly more formal proof.
Suppose towards contradiction that $varphicolon omega+1to omega$ is an isomorphism. Notice that $omegain omega+1$ is a maximal element.
Since order isomorphisms map maximal elements to maximal elements, it follows that $varphi(omega)$ is maximal in $omega$. But $omega$ has no maximal element, a contradiction, which finishes the proof.
As suggested by Henno Brandsma in the comments, this proof presupposes that we already know that being a maximal element is preserved by order isomorphism. If that is not completely clear, the proof of this fact should be included.
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
edited Mar 24 at 13:12
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
answered Mar 23 at 23:12
tomasztomasz
24.1k23482
24.1k23482
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
add a comment |
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
2
2
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
$begingroup$
Maybe the OP needs to prove the lemma that an order isomorphism maps maximal elements to maximal elements as well. Or it might have been proved before, it depends on how much his course covered so far and how formal his/her proofs are expected to be.
$endgroup$
– Henno Brandsma
Mar 24 at 6:34
add a comment |
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2
$begingroup$
What do you mean how do you write a proof? This pretty much is a proof. An isomorphism takes a maximal element to a maximal element. There is nothing to take $omega$ to, so there is no isomorphism.
$endgroup$
– tomasz
Mar 23 at 22:59
$begingroup$
@tomasz Yes but is this a mathematically proof?
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:00
2
$begingroup$
What do you mean by "mathematically proof"? In the sense of proof theory, not at all, but no one actually writes proofs that way. In the sense that it is a completely clear and sound reasoning, at least to me, yes. If it does not convince you, you should clarify your question with your doubts.
$endgroup$
– tomasz
Mar 23 at 23:07
$begingroup$
@tomasz okey, thanks
$endgroup$
– PozcuKushimotoStreet
Mar 23 at 23:09