Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is...












1












$begingroup$


I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.



My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.



I don't understand in which way the uncountable condition could solve the contradiction in the proof.










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  • $begingroup$
    With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
    $endgroup$
    – DanielWainfleet
    1 hour ago


















1












$begingroup$


I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.



My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.



I don't understand in which way the uncountable condition could solve the contradiction in the proof.










share|cite|improve this question











$endgroup$












  • $begingroup$
    With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
    $endgroup$
    – DanielWainfleet
    1 hour ago
















1












1








1





$begingroup$


I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.



My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.



I don't understand in which way the uncountable condition could solve the contradiction in the proof.










share|cite|improve this question











$endgroup$




I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.



My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.



I don't understand in which way the uncountable condition could solve the contradiction in the proof.







real-analysis analysis






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edited 3 hours ago







Tengerye

















asked 4 hours ago









TengeryeTengerye

1547




1547












  • $begingroup$
    With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
    $endgroup$
    – DanielWainfleet
    1 hour ago




















  • $begingroup$
    With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
    $endgroup$
    – DanielWainfleet
    1 hour ago


















$begingroup$
With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
$endgroup$
– DanielWainfleet
1 hour ago






$begingroup$
With the metric on $P$ inherited from the usual metric on $Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Ysubset X$ which is a bijective image of the set of all binary sequences.
$endgroup$
– DanielWainfleet
1 hour ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).



More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).



If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.



We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.






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$endgroup$













  • $begingroup$
    Thank you so much. I have revised my question.
    $endgroup$
    – Tengerye
    3 hours ago



















0












$begingroup$

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$



Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.



(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)



Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.






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$endgroup$













  • $begingroup$
    This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
    $endgroup$
    – DanielWainfleet
    1 hour ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).



More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).



If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.



We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much. I have revised my question.
    $endgroup$
    – Tengerye
    3 hours ago
















4












$begingroup$

First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).



More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).



If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.



We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much. I have revised my question.
    $endgroup$
    – Tengerye
    3 hours ago














4












4








4





$begingroup$

First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).



More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).



If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.



We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.






share|cite|improve this answer









$endgroup$



First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).



More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).



If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.



We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.







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answered 3 hours ago









Noah SchweberNoah Schweber

127k10151290




127k10151290












  • $begingroup$
    Thank you so much. I have revised my question.
    $endgroup$
    – Tengerye
    3 hours ago


















  • $begingroup$
    Thank you so much. I have revised my question.
    $endgroup$
    – Tengerye
    3 hours ago
















$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago




$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago











0












$begingroup$

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$



Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.



(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)



Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
    $endgroup$
    – DanielWainfleet
    1 hour ago


















0












$begingroup$

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$



Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.



(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)



Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
    $endgroup$
    – DanielWainfleet
    1 hour ago
















0












0








0





$begingroup$

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$



Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.



(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)



Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.






share|cite|improve this answer











$endgroup$



The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$



Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.



(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)



Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









DanielWainfleetDanielWainfleet

35.5k31648




35.5k31648












  • $begingroup$
    This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
    $endgroup$
    – DanielWainfleet
    1 hour ago




















  • $begingroup$
    This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
    $endgroup$
    – DanielWainfleet
    1 hour ago


















$begingroup$
This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
$endgroup$
– DanielWainfleet
1 hour ago






$begingroup$
This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set
$endgroup$
– DanielWainfleet
1 hour ago




















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