Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is...
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Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is uncountable.
Theorem 2.43 in Baby Rudin: How to understand the proof?Theorem 2.13 in Walter Rudin's Principles of Mathematical AnalysisWalter Rudin's Principle's of Mathematical AnalysisTheorem 2.43 in Baby Rudin: How to understand the proof?Trouble with Froda's Theorem Proof QuestionProof of Rudin's Theorem 2.43Understanding proof in Walter Rudin's Mathematical AnalysisProve that two sets A and B with $A cap B=emptyset$, $sup A = sup B$, $sup A notin A$ and $sup B notin B$ cannot exist.Let A be the set of all sequences of 0’s and 1’s (binary sequences). Prove that A is uncountable using Cantor's Diagonal Argument.Theorem 2.14 in Walter Rudin's Principles of Mathematical AnalysisWhy does this proof that the set of all finite subsets of N is a countable set not work for the set of all subsets of N?
$begingroup$
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.
real-analysis analysis
asked 4 hours ago
TengeryeTengerye
1547
$endgroup$
add a comment |
$begingroup$
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.
real-analysis analysis
asked 4 hours ago
TengeryeTengerye
1547
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
real-analysis analysis
asked 4 hours ago
TengeryeTengerye
1547
$endgroup$
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
real-analysis analysis
asked 4 hours ago
TengeryeTengerye
1547
asked 4 hours ago
TengeryeTengerye
1547
edited 3 hours ago
Tengerye
asked 4 hours ago
TengeryeTengerye
1547
asked 4 hours ago
TengeryeTengerye
1547
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago
add a comment |
$begingroup$
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.
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
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.
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
127k10151290
$begingroup$
Thank you so much. I have revised my question.
$endgroup$
– Tengerye
3 hours ago
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
$endgroup$
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.
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
edited 1 hour ago
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
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
add a comment |
$begingroup$
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
add a comment |
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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