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Question related to the use of the axiom of choice in real analysis, nonstandard analysis, and constructive...
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So, as far as I'm concerned, real analysis depends quite largely on some weak variants of the axiom of choice (such as the axiom of countable choice), and there seems to be no controversy surrounding this use of the axiom of choice. However, there has been significant criticism of nonstandard analysis (such as the notorious criticism by E. Bishop). My question is: do these criticisms come down to the fact that the construction of the hyperreal numbers depends on a lemma whose strength is on par with that of the axiom of choice (Zorn's lemma)? Or does it boil down to another reason?
Additionally, in response to Bishop's review, Keisler asks "why did Paul Halmos choose a constructivist as the review [of A. Robinson's Nonstandard Analysis]", which also stokes the question: how does the axiom of choice impede on constructive proofs?
I would much prefer answers in lay terms, but any help is equally appreciated. Thank you.
axiom-of-choice nonstandard-analysis
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
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show 4 more comments
$begingroup$
So, as far as I'm concerned, real analysis depends quite largely on some weak variants of the axiom of choice (such as the axiom of countable choice), and there seems to be no controversy surrounding this use of the axiom of choice. However, there has been significant criticism of nonstandard analysis (such as the notorious criticism by E. Bishop). My question is: do these criticisms come down to the fact that the construction of the hyperreal numbers depends on a lemma whose strength is on par with that of the axiom of choice (Zorn's lemma)? Or does it boil down to another reason?
Additionally, in response to Bishop's review, Keisler asks "why did Paul Halmos choose a constructivist as the review [of A. Robinson's Nonstandard Analysis]", which also stokes the question: how does the axiom of choice impede on constructive proofs?
I would much prefer answers in lay terms, but any help is equally appreciated. Thank you.
axiom-of-choice nonstandard-analysis
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
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Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
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– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32
|
show 4 more comments
$begingroup$
So, as far as I'm concerned, real analysis depends quite largely on some weak variants of the axiom of choice (such as the axiom of countable choice), and there seems to be no controversy surrounding this use of the axiom of choice. However, there has been significant criticism of nonstandard analysis (such as the notorious criticism by E. Bishop). My question is: do these criticisms come down to the fact that the construction of the hyperreal numbers depends on a lemma whose strength is on par with that of the axiom of choice (Zorn's lemma)? Or does it boil down to another reason?
Additionally, in response to Bishop's review, Keisler asks "why did Paul Halmos choose a constructivist as the review [of A. Robinson's Nonstandard Analysis]", which also stokes the question: how does the axiom of choice impede on constructive proofs?
I would much prefer answers in lay terms, but any help is equally appreciated. Thank you.
axiom-of-choice nonstandard-analysis
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
$endgroup$
So, as far as I'm concerned, real analysis depends quite largely on some weak variants of the axiom of choice (such as the axiom of countable choice), and there seems to be no controversy surrounding this use of the axiom of choice. However, there has been significant criticism of nonstandard analysis (such as the notorious criticism by E. Bishop). My question is: do these criticisms come down to the fact that the construction of the hyperreal numbers depends on a lemma whose strength is on par with that of the axiom of choice (Zorn's lemma)? Or does it boil down to another reason?
Additionally, in response to Bishop's review, Keisler asks "why did Paul Halmos choose a constructivist as the review [of A. Robinson's Nonstandard Analysis]", which also stokes the question: how does the axiom of choice impede on constructive proofs?
I would much prefer answers in lay terms, but any help is equally appreciated. Thank you.
axiom-of-choice nonstandard-analysis
axiom-of-choice nonstandard-analysis
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
asked Jan 16 at 9:58
joshuaheckroodtjoshuaheckroodt
1,205622
1,205622
$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
$begingroup$
Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32
|
show 4 more comments
$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
$begingroup$
Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32
$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
$begingroup$
Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32
|
show 4 more comments
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$begingroup$
Why do you need Zorn to build the hyperreals? You just need ultrafilters on $Bbb N$, this is quite weak, even just the ultrafilter lemma itself is significantly weaker than Zorn.. (You also need countable choice to ensure the proof of Łoś theorem works in this specific case.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:15
$begingroup$
Isn't it true that the proof that nonprincipal ultrafilters exist depends on Zorn's lemma? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:26
$begingroup$
Yes, "depend". But countable choice also depends on Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:26
$begingroup$
So how come NSA draws so much criticism in comparison to RA? @AsafKaragila
$endgroup$
– joshuaheckroodt
Jan 16 at 10:29
$begingroup$
Because it requires a different mindset, which may or may not be harder for people to learn. It also brings out the uses of choice in a much more pronounced way. Ultimately, if you want to do analysis, dependent choice and the ultrafilter lemma are probably your friend anyway. So the hyperreals are not that far away anymore.
$endgroup$
– Asaf Karagila♦
Jan 16 at 10:32