(ZF) Every nonempty perfect set in $mathbb{R}^k$ is uncountable.












1












$begingroup$


This is the part of proof in Rudin PMA p.41



Let $P(subset mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m in P$ such that (i) $overline {V_{n+1}}$ $subset V_n$ and (ii) $x_n notin overline {V_{n+1}}$ and (iii) $V_{n+1} cap P ≠ emptyset$. Then form a sequence ${V_n subset mathbb{R}^k | nin omega }$.



Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)



I want a proof without AC.
Help..






general-topology axiom-of-choice







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are you asking? Are you looking for some way to prove the result without choice?
    $endgroup$
    – KReiser
    Jul 30 '12 at 18:09










  • $begingroup$
    @KReiser Yes. I'll add it
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:11










  • $begingroup$
    You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:13










  • $begingroup$
    @Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:17












  • $begingroup$
    @Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:22
















1












$begingroup$


This is the part of proof in Rudin PMA p.41



Let $P(subset mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m in P$ such that (i) $overline {V_{n+1}}$ $subset V_n$ and (ii) $x_n notin overline {V_{n+1}}$ and (iii) $V_{n+1} cap P ≠ emptyset$. Then form a sequence ${V_n subset mathbb{R}^k | nin omega }$.



Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)



I want a proof without AC.
Help..






general-topology axiom-of-choice







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are you asking? Are you looking for some way to prove the result without choice?
    $endgroup$
    – KReiser
    Jul 30 '12 at 18:09










  • $begingroup$
    @KReiser Yes. I'll add it
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:11










  • $begingroup$
    You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:13










  • $begingroup$
    @Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:17












  • $begingroup$
    @Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:22














1












1








1


1



$begingroup$


This is the part of proof in Rudin PMA p.41



Let $P(subset mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m in P$ such that (i) $overline {V_{n+1}}$ $subset V_n$ and (ii) $x_n notin overline {V_{n+1}}$ and (iii) $V_{n+1} cap P ≠ emptyset$. Then form a sequence ${V_n subset mathbb{R}^k | nin omega }$.



Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)



I want a proof without AC.
Help..






general-topology axiom-of-choice







share|cite|improve this question











$endgroup$




This is the part of proof in Rudin PMA p.41



Let $P(subset mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m in P$ such that (i) $overline {V_{n+1}}$ $subset V_n$ and (ii) $x_n notin overline {V_{n+1}}$ and (iii) $V_{n+1} cap P ≠ emptyset$. Then form a sequence ${V_n subset mathbb{R}^k | nin omega }$.



Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)



I want a proof without AC.
Help..






general-topology axiom-of-choice




general-topology axiom-of-choice






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 30 '12 at 18:11







Katlus

















asked Jul 30 '12 at 18:02









KatlusKatlus

2,78521946




2,78521946












  • $begingroup$
    What are you asking? Are you looking for some way to prove the result without choice?
    $endgroup$
    – KReiser
    Jul 30 '12 at 18:09










  • $begingroup$
    @KReiser Yes. I'll add it
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:11










  • $begingroup$
    You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:13










  • $begingroup$
    @Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:17












  • $begingroup$
    @Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:22


















  • $begingroup$
    What are you asking? Are you looking for some way to prove the result without choice?
    $endgroup$
    – KReiser
    Jul 30 '12 at 18:09










  • $begingroup$
    @KReiser Yes. I'll add it
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:11










  • $begingroup$
    You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:13










  • $begingroup$
    @Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
    $endgroup$
    – Katlus
    Jul 30 '12 at 18:17












  • $begingroup$
    @Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 18:22
















$begingroup$
What are you asking? Are you looking for some way to prove the result without choice?
$endgroup$
– KReiser
Jul 30 '12 at 18:09




$begingroup$
What are you asking? Are you looking for some way to prove the result without choice?
$endgroup$
– KReiser
Jul 30 '12 at 18:09












$begingroup$
@KReiser Yes. I'll add it
$endgroup$
– Katlus
Jul 30 '12 at 18:11




$begingroup$
@KReiser Yes. I'll add it
$endgroup$
– Katlus
Jul 30 '12 at 18:11












$begingroup$
You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 18:13




$begingroup$
You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?)
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 18:13












$begingroup$
@Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
$endgroup$
– Katlus
Jul 30 '12 at 18:17






$begingroup$
@Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used..
$endgroup$
– Katlus
Jul 30 '12 at 18:17














$begingroup$
@Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 18:22




$begingroup$
@Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time.
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 18:22










1 Answer
1






active

oldest

votes


















3












$begingroup$

The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.



In the induction step, instead of taking "some $x_min P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.



Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.



Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't get the part 'rationals are well-ordered'
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:05










  • $begingroup$
    @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 19:07










  • $begingroup$
    Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:13










  • $begingroup$
    I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
    $endgroup$
    – YuiTo Cheng
    Dec 18 '18 at 15:02










  • $begingroup$
    @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 15:10











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.



In the induction step, instead of taking "some $x_min P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.



Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.



Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't get the part 'rationals are well-ordered'
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:05










  • $begingroup$
    @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 19:07










  • $begingroup$
    Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:13










  • $begingroup$
    I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
    $endgroup$
    – YuiTo Cheng
    Dec 18 '18 at 15:02










  • $begingroup$
    @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 15:10
















3












$begingroup$

The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.



In the induction step, instead of taking "some $x_min P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.



Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.



Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I don't get the part 'rationals are well-ordered'
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:05










  • $begingroup$
    @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 19:07










  • $begingroup$
    Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:13










  • $begingroup$
    I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
    $endgroup$
    – YuiTo Cheng
    Dec 18 '18 at 15:02










  • $begingroup$
    @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 15:10














3












3








3





$begingroup$

The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.



In the induction step, instead of taking "some $x_min P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.



Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.



Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.






share|cite|improve this answer









$endgroup$



The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.



In the induction step, instead of taking "some $x_min P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.



Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.



Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 30 '12 at 18:30









Asaf Karagila♦Asaf Karagila

305k33435765




305k33435765








  • 1




    $begingroup$
    I don't get the part 'rationals are well-ordered'
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:05










  • $begingroup$
    @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 19:07










  • $begingroup$
    Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:13










  • $begingroup$
    I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
    $endgroup$
    – YuiTo Cheng
    Dec 18 '18 at 15:02










  • $begingroup$
    @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 15:10














  • 1




    $begingroup$
    I don't get the part 'rationals are well-ordered'
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:05










  • $begingroup$
    @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
    $endgroup$
    – Asaf Karagila♦
    Jul 30 '12 at 19:07










  • $begingroup$
    Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
    $endgroup$
    – Katlus
    Jul 30 '12 at 19:13










  • $begingroup$
    I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
    $endgroup$
    – YuiTo Cheng
    Dec 18 '18 at 15:02










  • $begingroup$
    @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 15:10








1




1




$begingroup$
I don't get the part 'rationals are well-ordered'
$endgroup$
– Katlus
Jul 30 '12 at 19:05




$begingroup$
I don't get the part 'rationals are well-ordered'
$endgroup$
– Katlus
Jul 30 '12 at 19:05












$begingroup$
@Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 19:07




$begingroup$
@Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$.
$endgroup$
– Asaf Karagila♦
Jul 30 '12 at 19:07












$begingroup$
Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
$endgroup$
– Katlus
Jul 30 '12 at 19:13




$begingroup$
Oh got it, it's a different ordering of $mathbb{Q}$ from the usual. Thank you
$endgroup$
– Katlus
Jul 30 '12 at 19:13












$begingroup$
I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
$endgroup$
– YuiTo Cheng
Dec 18 '18 at 15:02




$begingroup$
I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC...
$endgroup$
– YuiTo Cheng
Dec 18 '18 at 15:02












$begingroup$
@YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 15:10




$begingroup$
@YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 15:10


















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