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Equivalence between Axiom of Choice, Zorn Lemma, Zermelo Theorem: doubt about a step of the proof.

1

$begingroup$

I am working through the mentioned proof for my bachelor thesis. My project is not strictly on logic or order theory, but rather on topological vector spaces: anyway, this fundamental theorem directly enters so many proofs I decided to include it.
I am studying Dugundji's book proof, which exploits f-towers. I find it very clear except from the following passage of which I post a photo. He is proooving Axiom Of Choice $Rightarrow$ Zorn's Lemma

I mean, shouldn't we check that $T_A$ are pairwise disjoint to apply AOC? It may be obvious but I can't see it right away.
Thanks in advance.

logic proof-explanation order-theory axiom-of-choice

share|cite|improve this question

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

$endgroup$

add a comment | 

1

$begingroup$

I am working through the mentioned proof for my bachelor thesis. My project is not strictly on logic or order theory, but rather on topological vector spaces: anyway, this fundamental theorem directly enters so many proofs I decided to include it.
I am studying Dugundji's book proof, which exploits f-towers. I find it very clear except from the following passage of which I post a photo. He is proooving Axiom Of Choice $Rightarrow$ Zorn's Lemma

I mean, shouldn't we check that $T_A$ are pairwise disjoint to apply AOC? It may be obvious but I can't see it right away.
Thanks in advance.

logic proof-explanation order-theory axiom-of-choice

share|cite|improve this question

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

$endgroup$

add a comment | 

$begingroup$

I am working through the mentioned proof for my bachelor thesis. My project is not strictly on logic or order theory, but rather on topological vector spaces: anyway, this fundamental theorem directly enters so many proofs I decided to include it.
I am studying Dugundji's book proof, which exploits f-towers. I find it very clear except from the following passage of which I post a photo. He is proooving Axiom Of Choice $Rightarrow$ Zorn's Lemma

I mean, shouldn't we check that $T_A$ are pairwise disjoint to apply AOC? It may be obvious but I can't see it right away.
Thanks in advance.

logic proof-explanation order-theory axiom-of-choice

share|cite|improve this question

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

$endgroup$

share|cite|improve this question

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

share|cite|improve this question

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

edited Mar 28 at 9:05

Asaf Karagila♦

307k33440773

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

asked Mar 28 at 8:52

Francesco BilottaFrancesco Bilotta

314

1 Answer
1

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oldest

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6

$begingroup$

The Axiom of Choice, many formulations it has.

The standard formulation is "If $X$ is a set of non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in x$ for all $xin X$."

But it turns out that often times you can twist the arms of this formulation to get something more amenable to whatever it is that you're trying to do at the moment. One common formulation is "If $X$ is a family of pairwise disjoint non-empty sets, then there is a set $T$ such that $|Tcap x|=1$ for all $xin X$", or "If $X$ is a family of pairwise disjoint non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in X$ for all $xin X$."

Now we need to show that these are equivalent. One direction is trivial, the other direction has been discussed on the site in length several times before. But the gist is that given $X$ you can define $Y=xtimes xmid xin X$, and that is a family of pairwise disjoint sets. Let $g$ be the choice function on $Y$, then $f(x)=operatornameproj_x(g(xtimes x))$ defines a choice function.

share|cite|improve this answer

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This definitely clears up my doubt: I was thinking the problem was much more restricted to the described situation! Thanks
  $endgroup$
  – Francesco Bilotta
  Mar 28 at 16:39
  

  
  
  
  

add a comment | 

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6

$begingroup$

The Axiom of Choice, many formulations it has.

The standard formulation is "If $X$ is a set of non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in x$ for all $xin X$."

But it turns out that often times you can twist the arms of this formulation to get something more amenable to whatever it is that you're trying to do at the moment. One common formulation is "If $X$ is a family of pairwise disjoint non-empty sets, then there is a set $T$ such that $|Tcap x|=1$ for all $xin X$", or "If $X$ is a family of pairwise disjoint non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in X$ for all $xin X$."

Now we need to show that these are equivalent. One direction is trivial, the other direction has been discussed on the site in length several times before. But the gist is that given $X$ you can define $Y=xtimes xmid xin X$, and that is a family of pairwise disjoint sets. Let $g$ be the choice function on $Y$, then $f(x)=operatornameproj_x(g(xtimes x))$ defines a choice function.

share|cite|improve this answer

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This definitely clears up my doubt: I was thinking the problem was much more restricted to the described situation! Thanks
  $endgroup$
  – Francesco Bilotta
  Mar 28 at 16:39
  

  
  
  
  

add a comment | 

6

$begingroup$

The Axiom of Choice, many formulations it has.

The standard formulation is "If $X$ is a set of non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in x$ for all $xin X$."

But it turns out that often times you can twist the arms of this formulation to get something more amenable to whatever it is that you're trying to do at the moment. One common formulation is "If $X$ is a family of pairwise disjoint non-empty sets, then there is a set $T$ such that $|Tcap x|=1$ for all $xin X$", or "If $X$ is a family of pairwise disjoint non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in X$ for all $xin X$."

Now we need to show that these are equivalent. One direction is trivial, the other direction has been discussed on the site in length several times before. But the gist is that given $X$ you can define $Y=xtimes xmid xin X$, and that is a family of pairwise disjoint sets. Let $g$ be the choice function on $Y$, then $f(x)=operatornameproj_x(g(xtimes x))$ defines a choice function.

share|cite|improve this answer

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This definitely clears up my doubt: I was thinking the problem was much more restricted to the described situation! Thanks
  $endgroup$
  – Francesco Bilotta
  Mar 28 at 16:39
  

  
  
  
  

add a comment | 

$begingroup$

The Axiom of Choice, many formulations it has.

The standard formulation is "If $X$ is a set of non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in x$ for all $xin X$."

But it turns out that often times you can twist the arms of this formulation to get something more amenable to whatever it is that you're trying to do at the moment. One common formulation is "If $X$ is a family of pairwise disjoint non-empty sets, then there is a set $T$ such that $|Tcap x|=1$ for all $xin X$", or "If $X$ is a family of pairwise disjoint non-empty sets, then there is a function $fcolon Xtobigcup X$ such that $f(x)in X$ for all $xin X$."

Now we need to show that these are equivalent. One direction is trivial, the other direction has been discussed on the site in length several times before. But the gist is that given $X$ you can define $Y=xtimes xmid xin X$, and that is a family of pairwise disjoint sets. Let $g$ be the choice function on $Y$, then $f(x)=operatornameproj_x(g(xtimes x))$ defines a choice function.

share|cite|improve this answer

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

$endgroup$

share|cite|improve this answer

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773

answered Mar 28 at 9:05

Asaf Karagila♦Asaf Karagila

307k33440773