Baire sets have arbitrarily fine refinements. Barry Simon. Problem 1 page 239.
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Definition (Baire set) Let X be a compact Hausdorff space. The Baire sets are the smallest $sigma$-algebra containing all compacts $G_{delta}$'s.
Definition (Partition) Given an algebra , $mathcal{U}$, a partition associated to $mathcal{U}$, is a finite subset $mathcal{P}subset mathcal{U}$ so that
(i) All sets in $mathcal{P}$ are nonempty
(ii) $P_1, P_2inmathcal{P}Rightarrow P_1cap P_2=emptyset$
(iii) $bigcup_{Pinmathcal{P}}P=X$
Problem:
Given any open cover $left{U_{a}right}_{ain I}$, of a compact Hausdorff space $X$, prove that one can find a partition $left{P_jright}_{j=1}^{n}$ into Baire sets so that each $P_j$ lies in some single $U_a$.
(Hint: First find an open cover by Baire sets, $left{V_lright}_{l=1}^{m}$, so each $V_l$ is in some $U_a$.
I have this...
Let $left{U_{a}right}_{ain I}$ open cover of $X$. $X$ is compact, then exists $1,ldots, m$ such that $left{U_{l}right}_{l=1}^{m}$ is finite open cover of $X$.
Now, $X=bigcup_{l=1}^{m} U_l$, and $bigcup_{l=1}^{m} U_l=bigcup_{l=1}^{m} U_lsetminus(U_1cup ldots, cup U_{l-1})$ with $U_{0}=emptyset.$ now this union is disjoint.
Define $P_l=U_lsetminus (U_1cup ldots cup U_{0})$. This works?...
general-topology compactness
asked Nov 17 at 23:44
eraldcoil
26119
add a comment |
up vote
1
down vote
favorite
Definition (Baire set) Let X be a compact Hausdorff space. The Baire sets are the smallest $sigma$-algebra containing all compacts $G_{delta}$'s.
Definition (Partition) Given an algebra , $mathcal{U}$, a partition associated to $mathcal{U}$, is a finite subset $mathcal{P}subset mathcal{U}$ so that
(i) All sets in $mathcal{P}$ are nonempty
(ii) $P_1, P_2inmathcal{P}Rightarrow P_1cap P_2=emptyset$
(iii) $bigcup_{Pinmathcal{P}}P=X$
Problem:
Given any open cover $left{U_{a}right}_{ain I}$, of a compact Hausdorff space $X$, prove that one can find a partition $left{P_jright}_{j=1}^{n}$ into Baire sets so that each $P_j$ lies in some single $U_a$.
(Hint: First find an open cover by Baire sets, $left{V_lright}_{l=1}^{m}$, so each $V_l$ is in some $U_a$.
I have this...
Let $left{U_{a}right}_{ain I}$ open cover of $X$. $X$ is compact, then exists $1,ldots, m$ such that $left{U_{l}right}_{l=1}^{m}$ is finite open cover of $X$.
Now, $X=bigcup_{l=1}^{m} U_l$, and $bigcup_{l=1}^{m} U_l=bigcup_{l=1}^{m} U_lsetminus(U_1cup ldots, cup U_{l-1})$ with $U_{0}=emptyset.$ now this union is disjoint.
Define $P_l=U_lsetminus (U_1cup ldots cup U_{0})$. This works?...
general-topology compactness
asked Nov 17 at 23:44
eraldcoil
26119
No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Definition (Baire set) Let X be a compact Hausdorff space. The Baire sets are the smallest $sigma$-algebra containing all compacts $G_{delta}$'s.
Definition (Partition) Given an algebra , $mathcal{U}$, a partition associated to $mathcal{U}$, is a finite subset $mathcal{P}subset mathcal{U}$ so that
(i) All sets in $mathcal{P}$ are nonempty
(ii) $P_1, P_2inmathcal{P}Rightarrow P_1cap P_2=emptyset$
(iii) $bigcup_{Pinmathcal{P}}P=X$
Problem:
Given any open cover $left{U_{a}right}_{ain I}$, of a compact Hausdorff space $X$, prove that one can find a partition $left{P_jright}_{j=1}^{n}$ into Baire sets so that each $P_j$ lies in some single $U_a$.
(Hint: First find an open cover by Baire sets, $left{V_lright}_{l=1}^{m}$, so each $V_l$ is in some $U_a$.
I have this...
Let $left{U_{a}right}_{ain I}$ open cover of $X$. $X$ is compact, then exists $1,ldots, m$ such that $left{U_{l}right}_{l=1}^{m}$ is finite open cover of $X$.
Now, $X=bigcup_{l=1}^{m} U_l$, and $bigcup_{l=1}^{m} U_l=bigcup_{l=1}^{m} U_lsetminus(U_1cup ldots, cup U_{l-1})$ with $U_{0}=emptyset.$ now this union is disjoint.
Define $P_l=U_lsetminus (U_1cup ldots cup U_{0})$. This works?...
general-topology compactness
asked Nov 17 at 23:44
eraldcoil
26119
Definition (Baire set) Let X be a compact Hausdorff space. The Baire sets are the smallest $sigma$-algebra containing all compacts $G_{delta}$'s.
Definition (Partition) Given an algebra , $mathcal{U}$, a partition associated to $mathcal{U}$, is a finite subset $mathcal{P}subset mathcal{U}$ so that
(i) All sets in $mathcal{P}$ are nonempty
(ii) $P_1, P_2inmathcal{P}Rightarrow P_1cap P_2=emptyset$
(iii) $bigcup_{Pinmathcal{P}}P=X$
Problem:
Given any open cover $left{U_{a}right}_{ain I}$, of a compact Hausdorff space $X$, prove that one can find a partition $left{P_jright}_{j=1}^{n}$ into Baire sets so that each $P_j$ lies in some single $U_a$.
(Hint: First find an open cover by Baire sets, $left{V_lright}_{l=1}^{m}$, so each $V_l$ is in some $U_a$.
I have this...
Let $left{U_{a}right}_{ain I}$ open cover of $X$. $X$ is compact, then exists $1,ldots, m$ such that $left{U_{l}right}_{l=1}^{m}$ is finite open cover of $X$.
Now, $X=bigcup_{l=1}^{m} U_l$, and $bigcup_{l=1}^{m} U_l=bigcup_{l=1}^{m} U_lsetminus(U_1cup ldots, cup U_{l-1})$ with $U_{0}=emptyset.$ now this union is disjoint.
Define $P_l=U_lsetminus (U_1cup ldots cup U_{0})$. This works?...
general-topology compactness
general-topology compactness
asked Nov 17 at 23:44
eraldcoil
26119
asked Nov 17 at 23:44
eraldcoil
26119
edited Nov 18 at 5:52
asked Nov 17 at 23:44
eraldcoil
26119
asked Nov 17 at 23:44
eraldcoil
26119
asked Nov 17 at 23:44
eraldcoil
26119
26119
No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20
add a comment |
No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20
No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20
add a comment |
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No, there is no reason why the $P_l$ you define are Baire sets. What have you learnt about Baire sets? They're Borel obviously. But do you have more info than that on such sets? Start looking there first.
– Henno Brandsma
Nov 18 at 6:15
I only know that the baire sets are in the smallest sigma algebra that contains the compacts formed by intersection of open sets.
– eraldcoil
Nov 18 at 15:27
$(U_a)$ open cover of X. for each $x$, exists $a_x$ such that $xin U_{a_(x)}$. ${x}$ is closed and $U_{a_x}$ open By Urysohn's lemma exists $f_x:Xto [0,1]$ ${f_x}(y)=1$ over ${x}$ and ${f_x}(y)=0$ over $Xsetminus U_{a_x}$ Then $left{y:f_x(y)>0right}subset U_{a_x}$ Now, ${y:{f_x}(y)>1/2}$ is open Baire set Then $V_x=left{y:{f_x}(y)>1/2right}$ is open cover Baire sets of $X$. And $X$ compact, then exists $x_1,ldots,x_m$ such that $X=bigcup_{l=1}^{m} V_l$ with $V_l=V_{x_l}$ Defines $P_l=V_{l}setminus (V_1 cupldots cup V_{l-1})$ $V_{0}=emptyset$ It is work?..
– eraldcoil
Nov 18 at 23:20