Baire's Category Theorem Proof
$begingroup$
I've searched online for this specific direction of proving Baire's Category Theorem but couldn't find anything substantive. I believe this attempts to prove it in the contrapositive way that, for example wikipedia, does. I'm posting it here to ask for proof verification. I'm following E.T. Copson, Metric Spaces.
Thrm: Any complete metric space $(X, d)$ cannot be expressed as a countable union of nowhere dense sets.
Proof: Sketch: We inductively constructed nested closed balls and use Cantor's Intersection Thrm to show that (with necessary conditions) $cap_n overline{B_{r_n}(x_n)} = {x}$ for some $x in X$, but that $x$ does not belong in any of the nowhere dense sets, thus contradiction.
Some facts:
1) Suppose it is, i.e. $X = cup_n S_n$ where $overline{Ext(S_n)} = X forall n in mathbb{N}$.
2) Then any $x in X$ is an adherent point, i.e. $forall epsilon > 0, B_{epsilon}(x) cap Ext(S_n) neq emptyset$.
3) Then for any $y in B_{epsilon}(x) cap Ext(S_n)$ for the $x$ in (2), we have $exists r > 0, overline{B_r(y)} subset Ext(S_n)$ (since $Ext(S_n)$ is open).
Now for construction:
Base: Construct $overline{B_{r_1}(x_1)}$ as follows: We know we can find some $overline{B_{r_1}(y_1)}$ such that it is disjoint from $S_1$. (from 2 and 3)) Denote this as $B_1$.
Induction: Given $B_{n-1} = cap_k^{n-1} overline{B_{r_k}(x_k)}$, we know $B$ is disjoint from $S_1, S_2, ..., S_{n-1}$. Now since we can find some $x in X$ that is an adherent point of $Ext(S_n)$ in $B$ (from 2), we can also find a closed ball $overline{B_{r_n}(y_n)}$ such that $r_n < r_{n-1}/2$(this is to make sure the diameter of the intersection tends to 0), and is disjoint from $S_n$ (from 3). Then we have $B_n = B_{n-1} cap overline{B_{r_n}(y_n)}$.
By our induction hypothesis and Cantor's intersection thrm, there must exist $x$ that is disjoint from all $S_n$'s, which is not possible because $X$ is constructed by a union of $S_n$'s.
real-analysis proof-verification metric-spaces baire-category
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
$endgroup$
add a comment |
$begingroup$
I've searched online for this specific direction of proving Baire's Category Theorem but couldn't find anything substantive. I believe this attempts to prove it in the contrapositive way that, for example wikipedia, does. I'm posting it here to ask for proof verification. I'm following E.T. Copson, Metric Spaces.
Thrm: Any complete metric space $(X, d)$ cannot be expressed as a countable union of nowhere dense sets.
Proof: Sketch: We inductively constructed nested closed balls and use Cantor's Intersection Thrm to show that (with necessary conditions) $cap_n overline{B_{r_n}(x_n)} = {x}$ for some $x in X$, but that $x$ does not belong in any of the nowhere dense sets, thus contradiction.
Some facts:
1) Suppose it is, i.e. $X = cup_n S_n$ where $overline{Ext(S_n)} = X forall n in mathbb{N}$.
2) Then any $x in X$ is an adherent point, i.e. $forall epsilon > 0, B_{epsilon}(x) cap Ext(S_n) neq emptyset$.
3) Then for any $y in B_{epsilon}(x) cap Ext(S_n)$ for the $x$ in (2), we have $exists r > 0, overline{B_r(y)} subset Ext(S_n)$ (since $Ext(S_n)$ is open).
Now for construction:
Base: Construct $overline{B_{r_1}(x_1)}$ as follows: We know we can find some $overline{B_{r_1}(y_1)}$ such that it is disjoint from $S_1$. (from 2 and 3)) Denote this as $B_1$.
Induction: Given $B_{n-1} = cap_k^{n-1} overline{B_{r_k}(x_k)}$, we know $B$ is disjoint from $S_1, S_2, ..., S_{n-1}$. Now since we can find some $x in X$ that is an adherent point of $Ext(S_n)$ in $B$ (from 2), we can also find a closed ball $overline{B_{r_n}(y_n)}$ such that $r_n < r_{n-1}/2$(this is to make sure the diameter of the intersection tends to 0), and is disjoint from $S_n$ (from 3). Then we have $B_n = B_{n-1} cap overline{B_{r_n}(y_n)}$.
By our induction hypothesis and Cantor's intersection thrm, there must exist $x$ that is disjoint from all $S_n$'s, which is not possible because $X$ is constructed by a union of $S_n$'s.
real-analysis proof-verification metric-spaces baire-category
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
$endgroup$
$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40
add a comment |
$begingroup$
I've searched online for this specific direction of proving Baire's Category Theorem but couldn't find anything substantive. I believe this attempts to prove it in the contrapositive way that, for example wikipedia, does. I'm posting it here to ask for proof verification. I'm following E.T. Copson, Metric Spaces.
Thrm: Any complete metric space $(X, d)$ cannot be expressed as a countable union of nowhere dense sets.
Proof: Sketch: We inductively constructed nested closed balls and use Cantor's Intersection Thrm to show that (with necessary conditions) $cap_n overline{B_{r_n}(x_n)} = {x}$ for some $x in X$, but that $x$ does not belong in any of the nowhere dense sets, thus contradiction.
Some facts:
1) Suppose it is, i.e. $X = cup_n S_n$ where $overline{Ext(S_n)} = X forall n in mathbb{N}$.
2) Then any $x in X$ is an adherent point, i.e. $forall epsilon > 0, B_{epsilon}(x) cap Ext(S_n) neq emptyset$.
3) Then for any $y in B_{epsilon}(x) cap Ext(S_n)$ for the $x$ in (2), we have $exists r > 0, overline{B_r(y)} subset Ext(S_n)$ (since $Ext(S_n)$ is open).
Now for construction:
Base: Construct $overline{B_{r_1}(x_1)}$ as follows: We know we can find some $overline{B_{r_1}(y_1)}$ such that it is disjoint from $S_1$. (from 2 and 3)) Denote this as $B_1$.
Induction: Given $B_{n-1} = cap_k^{n-1} overline{B_{r_k}(x_k)}$, we know $B$ is disjoint from $S_1, S_2, ..., S_{n-1}$. Now since we can find some $x in X$ that is an adherent point of $Ext(S_n)$ in $B$ (from 2), we can also find a closed ball $overline{B_{r_n}(y_n)}$ such that $r_n < r_{n-1}/2$(this is to make sure the diameter of the intersection tends to 0), and is disjoint from $S_n$ (from 3). Then we have $B_n = B_{n-1} cap overline{B_{r_n}(y_n)}$.
By our induction hypothesis and Cantor's intersection thrm, there must exist $x$ that is disjoint from all $S_n$'s, which is not possible because $X$ is constructed by a union of $S_n$'s.
real-analysis proof-verification metric-spaces baire-category
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
$endgroup$
I've searched online for this specific direction of proving Baire's Category Theorem but couldn't find anything substantive. I believe this attempts to prove it in the contrapositive way that, for example wikipedia, does. I'm posting it here to ask for proof verification. I'm following E.T. Copson, Metric Spaces.
Thrm: Any complete metric space $(X, d)$ cannot be expressed as a countable union of nowhere dense sets.
Proof: Sketch: We inductively constructed nested closed balls and use Cantor's Intersection Thrm to show that (with necessary conditions) $cap_n overline{B_{r_n}(x_n)} = {x}$ for some $x in X$, but that $x$ does not belong in any of the nowhere dense sets, thus contradiction.
Some facts:
1) Suppose it is, i.e. $X = cup_n S_n$ where $overline{Ext(S_n)} = X forall n in mathbb{N}$.
2) Then any $x in X$ is an adherent point, i.e. $forall epsilon > 0, B_{epsilon}(x) cap Ext(S_n) neq emptyset$.
3) Then for any $y in B_{epsilon}(x) cap Ext(S_n)$ for the $x$ in (2), we have $exists r > 0, overline{B_r(y)} subset Ext(S_n)$ (since $Ext(S_n)$ is open).
Now for construction:
Base: Construct $overline{B_{r_1}(x_1)}$ as follows: We know we can find some $overline{B_{r_1}(y_1)}$ such that it is disjoint from $S_1$. (from 2 and 3)) Denote this as $B_1$.
Induction: Given $B_{n-1} = cap_k^{n-1} overline{B_{r_k}(x_k)}$, we know $B$ is disjoint from $S_1, S_2, ..., S_{n-1}$. Now since we can find some $x in X$ that is an adherent point of $Ext(S_n)$ in $B$ (from 2), we can also find a closed ball $overline{B_{r_n}(y_n)}$ such that $r_n < r_{n-1}/2$(this is to make sure the diameter of the intersection tends to 0), and is disjoint from $S_n$ (from 3). Then we have $B_n = B_{n-1} cap overline{B_{r_n}(y_n)}$.
By our induction hypothesis and Cantor's intersection thrm, there must exist $x$ that is disjoint from all $S_n$'s, which is not possible because $X$ is constructed by a union of $S_n$'s.
real-analysis proof-verification metric-spaces baire-category
real-analysis proof-verification metric-spaces baire-category
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
edited Dec 6 '18 at 3:55
Alex Ravsky
40.4k32282
40.4k32282
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 0:36
OneRaynyDayOneRaynyDay
306111
306111
$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40
add a comment |
$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40
$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40
add a comment |
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$begingroup$
This is off topic in the sense that it is unclear what you are asking for. The Wikipedia page gives a proof and presumably Copson gives a proof. What are you asking us to do?
$endgroup$
– Rob Arthan
Nov 28 '18 at 1:28
$begingroup$
If I’m not mistaken, the Wikipedia page presents 3 different approaches to the baires category theorem, and only BCT1 is given a proof on the main page. The one I am trying to prove is BCT3, which is not on the page. One can make an argument that BCT3 follows from BCT1 but I’m trying to prove it directly.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:33
$begingroup$
I forgot to note that I’m not 100% sure if this is the same as Copsons proof, and I’m asking for verification that it indeed is.
$endgroup$
– OneRaynyDay
Nov 28 '18 at 1:40