Help with AC $Rightarrow$ Zorn's Lemma proof
$begingroup$
I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf
Specifically the last section of this part:
Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)
Thanks in advance
logic set-theory axiom-of-choice
asked Dec 6 '18 at 1:38
mate89mate89
1799
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add a comment |
$begingroup$
I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf
Specifically the last section of this part:
Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)
Thanks in advance
logic set-theory axiom-of-choice
asked Dec 6 '18 at 1:38
mate89mate89
1799
$endgroup$
add a comment |
$begingroup$
I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf
Specifically the last section of this part:
Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)
Thanks in advance
logic set-theory axiom-of-choice
asked Dec 6 '18 at 1:38
mate89mate89
1799
$endgroup$
I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf
Specifically the last section of this part:
Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)
Thanks in advance
logic set-theory axiom-of-choice
logic set-theory axiom-of-choice
asked Dec 6 '18 at 1:38
mate89mate89
1799
asked Dec 6 '18 at 1:38
mate89mate89
1799
edited Dec 6 '18 at 2:03
mate89
asked Dec 6 '18 at 1:38
mate89mate89
1799
asked Dec 6 '18 at 1:38
mate89mate89
1799
asked Dec 6 '18 at 1:38
mate89mate89
1799
1799
add a comment |
add a comment |
1 Answer
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Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).
And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)
Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).
And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)
Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
add a comment |
$begingroup$
Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).
And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)
Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
add a comment |
$begingroup$
Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).
And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)
Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).
And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)
Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
answered Dec 6 '18 at 1:49
Alex KruckmanAlex Kruckman
27.2k22557
27.2k22557
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
add a comment |
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
$begingroup$
You sir, you just saved the day. Thanks a lot.
$endgroup$
– mate89
Dec 6 '18 at 1:59
add a comment |
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