proof that axiom of union and axiom of pairing imply weak axiom of union

The axiom of union says that to every set $mathcal{A}$ there is a set ${z: exists A ; z in A wedge A in mathcal{A}}$ or in other words: $forall mathcal{A} exists B: forall z [z in B leftrightarrow exists A in mathcal{A} ; z in A]$.

The weak axiom of union says that $forall A forall B exists C: forall x (x in C leftrightarrow x in A vee x in B)$.

The axiom of pairing says: $forall x forall y exists p: forall z [z in p leftrightarrow z = x vee z = y]$.

I am trying to give a (not fully formal) proof that if the axiom of union and axiom of pairing do hold, then also the weak axiom of pairing does hold.

Futhermore, I'm looking for a structure in which the axiom of union does hold, but the weak axiom of union doesn't. (So obviously this has to be a structure where the axiom of union does hold, but the axiom of pairing does not.)
I'm still trying to find an apt structure, but I think that finally I'll find one.

So I'd primarily appreciate your help on giving a (nearly formal) proof, since I'm just not managing to. Thank you in advance!

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

add a comment |

I am trying to give a (not fully formal) proof that if the axiom of union and axiom of pairing do hold, then also the weak axiom of pairing does hold.

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

add a comment |

I am trying to give a (not fully formal) proof that if the axiom of union and axiom of pairing do hold, then also the weak axiom of pairing does hold.

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

I am trying to give a (not fully formal) proof that if the axiom of union and axiom of pairing do hold, then also the weak axiom of pairing does hold.

elementary-set-theory logic proof-writing

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

edited Jan 12 at 18:42

Andrés E. Caicedo

65.7k8160250

asked Jan 12 at 15:48

Studentu

1279

asked Jan 12 at 15:48

Studentu

1279

asked Jan 12 at 15:48

Studentu

1279

add a comment |

2 Answers
2

active

oldest

votes

Assume the axiom of union and the axiom of pairing hold. Then by the axiom of pairing $$forall Aforall Bexists p:forall S[Sin piff S=Avee S=B],$$ and by the axiom of union $$forall pexists C:forall z[zin Ciffexists Sin p:zin S].$$ Therefore $$forall Aforall Bexists C:forall z[zin Ciffexists S:(S=Avee S=B)wedge zin S].$$ I will leave the details to you.

EDIT: Full proof: Let $A,B$ be sets. By the axiom of pairing there exists a set $p$ such that $$forall S[Sin piff S=Avee S=B].$$ It follows that $forall z:$
begin{align}
(exists Sin p:zin S)
&iff(exists S:(S=Avee S=B)wedge zin S)
\&iff(exists S:(S=Awedge zin S)vee(S=Bwedge zin S))
\&iff(zin Avee zin B).
end{align}
By the axiom of union there exists a set $C$ such that $$forall z[zin Ciffexists Sin p:zin S].$$ Hence, we find $$forall z[zin Ciff zin Avee zin B].$$ Since $A,B$ were arbitrary, we conclude $$forall Aforall Bexists C:forall z[zin Ciff zin Avee zin B].$$ So if the axiom of pairing and the axiom of union hold, then the weak axiom of union holds.

Fixed counterexample: A model where the axiom of union holds but the weak axiom of union does not hold would be the model where $emptyset$, ${emptyset}$ and ${{emptyset}}$ are the only sets. The weak axiom of union would imply that ${emptyset}cup{{emptyset}}={emptyset,{emptyset}}$ would exist, which it does not. I leave it up to you to check that the axiom of union does hold.

edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

$begingroup$
Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?
$endgroup$
– Studentu
Jan 12 at 17:04

1

$begingroup$
I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.
$endgroup$
– SmileyCraft
Jan 12 at 18:17

$begingroup$
Cool, thank you, SmileyCraft!
$endgroup$
– Studentu
Jan 13 at 16:59

add a comment |

For any sets $A$ and $B$ there is a set $C = { A, B }$. By Union, there must be a set $D$ that contains all and only the elements of the elements of $C$, i.e. $D$ will contain all and only elements of either $A$ or $B$. That's the set $'C'$ you are looking for in the definition of Weak Union .. not the initial $C$

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof:

enter image description here

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

$begingroup$
Thanks for your help, Bram28!
$endgroup$
– Studentu
Jan 13 at 18:08

1

$begingroup$
@Studentu You're welcome :)
$endgroup$
– Bram28
Jan 13 at 18:18

add a comment |

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2 Answers
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2 Answers
2

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oldest

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edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

$begingroup$
Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?
$endgroup$
– Studentu
Jan 12 at 17:04

1

$begingroup$
I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.
$endgroup$
– SmileyCraft
Jan 12 at 18:17

$begingroup$
Cool, thank you, SmileyCraft!
$endgroup$
– Studentu
Jan 13 at 16:59

add a comment |

edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

$begingroup$
Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?
$endgroup$
– Studentu
Jan 12 at 17:04

1

$begingroup$
I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.
$endgroup$
– SmileyCraft
Jan 12 at 18:17

$begingroup$
Cool, thank you, SmileyCraft!
$endgroup$
– Studentu
Jan 13 at 16:59

add a comment |

edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

edited Jan 12 at 18:17

answered Jan 12 at 16:03

SmileyCraft

3,709519

answered Jan 12 at 16:03

SmileyCraft

3,709519

answered Jan 12 at 16:03

SmileyCraft

3,709519

$begingroup$
Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?
$endgroup$
– Studentu
Jan 12 at 17:04

1

$begingroup$
I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.
$endgroup$
– SmileyCraft
Jan 12 at 18:17

$begingroup$
Cool, thank you, SmileyCraft!
$endgroup$
– Studentu
Jan 13 at 16:59

add a comment |

$begingroup$
Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?
$endgroup$
– Studentu
Jan 12 at 16:37

$begingroup$
And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?
$endgroup$
– Studentu
Jan 12 at 17:04

1

$begingroup$
I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.
$endgroup$
– SmileyCraft
Jan 12 at 18:17

$begingroup$
Cool, thank you, SmileyCraft!
$endgroup$
– Studentu
Jan 13 at 16:59

Thank you for your fast reply, SmileyCraft! But it are the details I am struggling with. Could you show me either how to proof that ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S] follows from ∀A∀B∃p:∀S[S∈p⇔S=A∨S=B]∧∀Ap∃C:∀z[z∈C⇔∃S$in p:$z∈S]

– Studentu
Jan 12 at 16:37

or how to proof ∀A∀B∃C:∀z[z∈C⇔z∈A∨z∈B] from ∀A∀B∃C:∀z[z∈C⇔∃S:(S=A∨S=B)∧z∈S]?

– Studentu
Jan 12 at 16:37

And I don't see how your example doesn't fulfill the weak axiom of union: 1) $A = emptyset, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. $A = emptyset, B = emptyset$, then $C = emptyset$ fulfills what we want. $A = {emptyset}, B = {emptyset}$, then $C = {emptyset}$ fulfills what we want. So we don't get any contradiction to the weak axiom of union?

– Studentu
Jan 12 at 17:04

I added a full fool proof proof :) And you were correct that my example was flawed, I fixed it.

– SmileyCraft
Jan 12 at 18:17

Cool, thank you, SmileyCraft!

– Studentu
Jan 13 at 16:59

add a comment |

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof:

enter image description here

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

$begingroup$
Thanks for your help, Bram28!
$endgroup$
– Studentu
Jan 13 at 18:08

1

$begingroup$
@Studentu You're welcome :)
$endgroup$
– Bram28
Jan 13 at 18:18

add a comment |

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof:

enter image description here

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

$begingroup$
Thanks for your help, Bram28!
$endgroup$
– Studentu
Jan 13 at 18:08

1

$begingroup$
@Studentu You're welcome :)
$endgroup$
– Bram28
Jan 13 at 18:18

add a comment |

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof:

enter image description here

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof:

enter image description here

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

edited Jan 12 at 18:27

answered Jan 12 at 18:18

Bram28

63.7k44793

answered Jan 12 at 18:18

Bram28

63.7k44793

answered Jan 12 at 18:18

Bram28

63.7k44793

$begingroup$
Thanks for your help, Bram28!
$endgroup$
– Studentu
Jan 13 at 18:08

1

$begingroup$
@Studentu You're welcome :)
$endgroup$
– Bram28
Jan 13 at 18:18

add a comment |

$begingroup$
Thanks for your help, Bram28!
$endgroup$
– Studentu
Jan 13 at 18:08

1

$begingroup$
@Studentu You're welcome :)
$endgroup$
– Bram28
Jan 13 at 18:18

Thanks for your help, Bram28!

– Studentu
Jan 13 at 18:08

@Studentu You're welcome :)

– Bram28
Jan 13 at 18:18

add a comment |

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