How do I prove the anti-symmetry and there is a minimal element for each subset of $alpha$Prove using transfinite induction that if ordinals $alpha$ and $beta$ are countable, then so is $alpha + beta$.Element of ordinal a subset of the same ordinalQuestion about the proof of this lemma: If $alpha$, $beta$ are ordinals, then either $alpha subset beta$ or $beta subset alpha.$For every ordinal $alpha$, there is a cardinal number greater then $alpha$.Prove divisibility is a partial order relation over natural numbersTrichotomy of Ordinals. Is $K$ a set?Transitivity and anti-symmetry of setHow to prove that a subset of an oridnal is an element of the ordinal by transitivity.Ordinal numbers, the Burali-Forti paradox, and anti-foundation axiomsI'm trying to prove that any finite partially ordered set has a minimal element.
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How do I prove the anti-symmetry and there is a minimal element for each subset of $alpha$
Prove using transfinite induction that if ordinals $alpha$ and $beta$ are countable, then so is $alpha + beta$.Element of ordinal a subset of the same ordinalQuestion about the proof of this lemma: If $alpha$, $beta$ are ordinals, then either $alpha subset beta$ or $beta subset alpha.$For every ordinal $alpha$, there is a cardinal number greater then $alpha$.Prove divisibility is a partial order relation over natural numbersTrichotomy of Ordinals. Is $K$ a set?Transitivity and anti-symmetry of setHow to prove that a subset of an oridnal is an element of the ordinal by transitivity.Ordinal numbers, the Burali-Forti paradox, and anti-foundation axiomsI'm trying to prove that any finite partially ordered set has a minimal element.
$begingroup$
An ordinal number is a set $alpha$ with the following properties:
(a) If $x,y in alpha,$ then either $xin y$, $yin x$, or $x=y.$
(b)If $yin alpha$ and $xin y$, then $xin alpha$.
Let $alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $alpha$, define $leq$ on $alpha$ by $xleq y$ iff $x=y$ or $xin y$.
Theorem:-
Let $alpha$ is an ordinal number. Then $(alpha,leq)$ is a well ordered set.
Proof:-
I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $alpha$.
Doubt 1. anti-symmetric
$xleq y$ and $yleq x$. Our claim is $x=y.$
Case 1 $xleq y$ and $yleq x$ means $x=y$ and $yin x implies xin x$. Which is a paradox. How do I prove the anti-symmetry?
Case 2 $xleq y$ and $yleq x$ means $x=y$ and $y= x implies x= y$.
Case 3 $xleq y$ and $yleq x$ means $xin y$ and $y= x implies xin x$.
Case 4 $xleq y$ and $yleq x$ means $xin y$ and $yin x implies x= y$.
Also How do I prove there is a minimal elemnt for each subset of $alpha$?
elementary-set-theory ordinals
$endgroup$
add a comment |
$begingroup$
An ordinal number is a set $alpha$ with the following properties:
(a) If $x,y in alpha,$ then either $xin y$, $yin x$, or $x=y.$
(b)If $yin alpha$ and $xin y$, then $xin alpha$.
Let $alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $alpha$, define $leq$ on $alpha$ by $xleq y$ iff $x=y$ or $xin y$.
Theorem:-
Let $alpha$ is an ordinal number. Then $(alpha,leq)$ is a well ordered set.
Proof:-
I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $alpha$.
Doubt 1. anti-symmetric
$xleq y$ and $yleq x$. Our claim is $x=y.$
Case 1 $xleq y$ and $yleq x$ means $x=y$ and $yin x implies xin x$. Which is a paradox. How do I prove the anti-symmetry?
Case 2 $xleq y$ and $yleq x$ means $x=y$ and $y= x implies x= y$.
Case 3 $xleq y$ and $yleq x$ means $xin y$ and $y= x implies xin x$.
Case 4 $xleq y$ and $yleq x$ means $xin y$ and $yin x implies x= y$.
Also How do I prove there is a minimal elemnt for each subset of $alpha$?
elementary-set-theory ordinals
$endgroup$
add a comment |
$begingroup$
An ordinal number is a set $alpha$ with the following properties:
(a) If $x,y in alpha,$ then either $xin y$, $yin x$, or $x=y.$
(b)If $yin alpha$ and $xin y$, then $xin alpha$.
Let $alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $alpha$, define $leq$ on $alpha$ by $xleq y$ iff $x=y$ or $xin y$.
Theorem:-
Let $alpha$ is an ordinal number. Then $(alpha,leq)$ is a well ordered set.
Proof:-
I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $alpha$.
Doubt 1. anti-symmetric
$xleq y$ and $yleq x$. Our claim is $x=y.$
Case 1 $xleq y$ and $yleq x$ means $x=y$ and $yin x implies xin x$. Which is a paradox. How do I prove the anti-symmetry?
Case 2 $xleq y$ and $yleq x$ means $x=y$ and $y= x implies x= y$.
Case 3 $xleq y$ and $yleq x$ means $xin y$ and $y= x implies xin x$.
Case 4 $xleq y$ and $yleq x$ means $xin y$ and $yin x implies x= y$.
Also How do I prove there is a minimal elemnt for each subset of $alpha$?
elementary-set-theory ordinals
$endgroup$
An ordinal number is a set $alpha$ with the following properties:
(a) If $x,y in alpha,$ then either $xin y$, $yin x$, or $x=y.$
(b)If $yin alpha$ and $xin y$, then $xin alpha$.
Let $alpha$ be an ordinal number. For any two numbers $x$ and $y$ of $alpha$, define $leq$ on $alpha$ by $xleq y$ iff $x=y$ or $xin y$.
Theorem:-
Let $alpha$ is an ordinal number. Then $(alpha,leq)$ is a well ordered set.
Proof:-
I don't have the doubts in the proof of Reflexivity, Transitivity. I could be able to prove each element is comparable in $alpha$.
Doubt 1. anti-symmetric
$xleq y$ and $yleq x$. Our claim is $x=y.$
Case 1 $xleq y$ and $yleq x$ means $x=y$ and $yin x implies xin x$. Which is a paradox. How do I prove the anti-symmetry?
Case 2 $xleq y$ and $yleq x$ means $x=y$ and $y= x implies x= y$.
Case 3 $xleq y$ and $yleq x$ means $xin y$ and $y= x implies xin x$.
Case 4 $xleq y$ and $yleq x$ means $xin y$ and $yin x implies x= y$.
Also How do I prove there is a minimal elemnt for each subset of $alpha$?
elementary-set-theory ordinals
elementary-set-theory ordinals
edited Mar 17 at 12:43
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Mar 17 at 11:59
Math geekMath geek
50111
50111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.
$endgroup$
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
add a comment |
$begingroup$
I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $leq$ is an anti-symmetric relation on $alpha$.
Let $A$ be a non-empty subset of $alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.
So if $xin A$ then we do not have: $xin a$.
But what we do have is: $xin avee x=avee ain x$.
So what remains is: $x=avee ain x$ or equivalently $aleq x$.
This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
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votes
active
oldest
votes
active
oldest
votes
$begingroup$
Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.
$endgroup$
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
add a comment |
$begingroup$
Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.
$endgroup$
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
add a comment |
$begingroup$
Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.
$endgroup$
Anti-symmetry may be proven the following way: Let $x, yin alpha$ such that $xleq y$ and $yleq x$. This means that we have
$$
xin y text or x = y\
textand\
yin x text or y = x
$$
So, one possibility is certainly that $x = y$. If $xneq y$, then that means that we must have $xin y$ and $yin x$, which cannot be true (as the set $x, y$ would violate the axiom of foundation / regularity). This proves anti-symmetry.
As for well-orderedness, take a non-empty subset $Ssubseteq alpha$. By the axiom of foundation, there is an element $xin S$ such that $xcap S = varnothing$. I claim that this $x$ is minimal in $S$.
Assume, for contradiction, that $x$ is not minimal. Then there is a $yin S$ with $yleq x, yneq x$. This means that $yin x$. But then we have $yin xcap S$, which is a contradiction.
answered Mar 17 at 12:16
ArthurArthur
120k7120203
120k7120203
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
add a comment |
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
$begingroup$
You mean axiom of foundation as Russel's paradox?
$endgroup$
– Math geek
Mar 17 at 12:28
2
2
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
$begingroup$
@Mathgeek What do you mean? If you want to argue about sets on a fundamental level, you have to be aware of your axiomatic system. I assumed the ZF axioms, because that's the most common one. The axiom of foundation is one of the ZF axioms, and is specifically needed to avoid sets like $xin x$, or as in my case, $xin y, yin x$. You can do set theory with the ZF axioms except foundation, and in that case $xin x$ is entirely possible and not paradoxical at all. Particularily, there is no Russel's paradox in sight anywhere.
$endgroup$
– Arthur
Mar 17 at 12:36
add a comment |
$begingroup$
I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $leq$ is an anti-symmetric relation on $alpha$.
Let $A$ be a non-empty subset of $alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.
So if $xin A$ then we do not have: $xin a$.
But what we do have is: $xin avee x=avee ain x$.
So what remains is: $x=avee ain x$ or equivalently $aleq x$.
This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.
$endgroup$
add a comment |
$begingroup$
I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $leq$ is an anti-symmetric relation on $alpha$.
Let $A$ be a non-empty subset of $alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.
So if $xin A$ then we do not have: $xin a$.
But what we do have is: $xin avee x=avee ain x$.
So what remains is: $x=avee ain x$ or equivalently $aleq x$.
This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.
$endgroup$
add a comment |
$begingroup$
I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $leq$ is an anti-symmetric relation on $alpha$.
Let $A$ be a non-empty subset of $alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.
So if $xin A$ then we do not have: $xin a$.
But what we do have is: $xin avee x=avee ain x$.
So what remains is: $x=avee ain x$ or equivalently $aleq x$.
This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.
$endgroup$
I preassume that the axiom of regularity is accepted in this context.
Then consequently it cannot happen that $xin x$ so the cases 1 and 3 will not show up.
Also the axiom excludes the situation that $xin yin x$ so also case 4 is excluded.
Our conclusion is then that we are dealing with case 2 as was to be shown.
Proved is now that $leq$ is an anti-symmetric relation on $alpha$.
Let $A$ be a non-empty subset of $alpha$.
According to the axiom this set contains an element $a$ such that all elements of $a$ are not elements of $A$.
So if $xin A$ then we do not have: $xin a$.
But what we do have is: $xin avee x=avee ain x$.
So what remains is: $x=avee ain x$ or equivalently $aleq x$.
This proves that $a$ serves as least element of $A$ in the order $(alpha,leq)$.
answered Mar 17 at 12:29
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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