Mixing
Problem 2. A comprehensive course in Analysis. Barry simon. Page 239.
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) $bigcup_{Pinmathcal{P}}P=X$
Given any continuos function $f$, on a compact Hausdorff space and any $epsilon>0$, find a Baire partition $left{P_jright}_{j=1}^{n}$ so that $sup_{x,yin P_j} |f(x)-f(y)|<epsilon$.
Hint: First find an open cover by Baire sets, $left{U_lright}_{l=1}^{n}$ so that for each $l, sup_{x,yin P_j}|f(x)-f(y)|<epsilon.$
I do not know how to do this problem. Some help.?
functional-analysis measure-theory compactness descriptive-set-theory baire-category
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
add a comment |
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) $bigcup_{Pinmathcal{P}}P=X$
Given any continuos function $f$, on a compact Hausdorff space and any $epsilon>0$, find a Baire partition $left{P_jright}_{j=1}^{n}$ so that $sup_{x,yin P_j} |f(x)-f(y)|<epsilon$.
Hint: First find an open cover by Baire sets, $left{U_lright}_{l=1}^{n}$ so that for each $l, sup_{x,yin P_j}|f(x)-f(y)|<epsilon.$
I do not know how to do this problem. Some help.?
functional-analysis measure-theory compactness descriptive-set-theory baire-category
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15
add a comment |
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) $bigcup_{Pinmathcal{P}}P=X$
Given any continuos function $f$, on a compact Hausdorff space and any $epsilon>0$, find a Baire partition $left{P_jright}_{j=1}^{n}$ so that $sup_{x,yin P_j} |f(x)-f(y)|<epsilon$.
Hint: First find an open cover by Baire sets, $left{U_lright}_{l=1}^{n}$ so that for each $l, sup_{x,yin P_j}|f(x)-f(y)|<epsilon.$
I do not know how to do this problem. Some help.?
functional-analysis measure-theory compactness descriptive-set-theory baire-category
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
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) $bigcup_{Pinmathcal{P}}P=X$
Given any continuos function $f$, on a compact Hausdorff space and any $epsilon>0$, find a Baire partition $left{P_jright}_{j=1}^{n}$ so that $sup_{x,yin P_j} |f(x)-f(y)|<epsilon$.
Hint: First find an open cover by Baire sets, $left{U_lright}_{l=1}^{n}$ so that for each $l, sup_{x,yin P_j}|f(x)-f(y)|<epsilon.$
I do not know how to do this problem. Some help.?
functional-analysis measure-theory compactness descriptive-set-theory baire-category
functional-analysis measure-theory compactness descriptive-set-theory baire-category
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
edited Nov 18 '18 at 2:22
Nate Eldredge
62.4k682170
62.4k682170
asked Nov 18 '18 at 0:38
eraldcoil
363111
asked Nov 18 '18 at 0:38
eraldcoil
363111
asked Nov 18 '18 at 0:38
eraldcoil
363111
363111
Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15
add a comment |
Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15
Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15
add a comment |
1 Answer
1
active
oldest
votes
Using the hint, our plan is to first find a finite open cover, followed by a finite compact $G_delta$ cover and finally a Baire partition.
Let $varepsilon > 0$ and $x in X$. By continuity of $f$, we can find an open $U_x$ containing $x$ such that for all $u in U_x$ we have
$$|f(x) - f(u)| < frac{varepsilon}{4}$$
This implies for all $y, z in U_x$ we get
$$begin{align}
|f(y) - f(z)| &le |f(y) - f(x)| + |f(x) - f(z)| \
&< frac{varepsilon}{4} + frac{varepsilon}{4} \
&= frac{varepsilon}{2}
end{align}$$
By order preserving property of supremum, we obtain
$$sup_{y, z in U_x} |f(y) - f(z)| le frac{varepsilon}{2} < varepsilon$$
Next, since $X$ is locally compact Hausdorff, we can choose a compact neighbourhood $K_x$ and open $V_x$ such that $$x in V_x subseteq K_x subseteq U_x$$
Now by lemma 3 of this question (please check!),
we can pick a compact $G_delta$ set $G_x$ such that $$x in V_x subseteq K_x subseteq G_x subseteq U_x$$
Observe that ${V_x}_{x in X}$ is an open cover for $X$.
By compactness of $X$, there is a finite subcover ${V_{x_1}, dots, V_{x_n}}$. Since each $V_{x_j} subseteq G_{x_j}$, ${G_{x_1}, dots, G_{x_n}}$ is also a finite cover. Hence ${G_{x_1}, dots, G_{x_n}}$ is a finite compact $G_delta$ cover. Besides, since each $G_{x_j} subseteq U_{x_j}$, we have $$sup_{y, z in G_{x_j}} |f(y) - f(z)| < varepsilon$$
Finally, let $$begin{align}
mathcal{P} = {&B_1 cap dots cap B_n mid \
&B_j = G_{x_j} text{ or } X setminus G_{x_j} text{ with } B_1 cap dots cap B_n neq emptyset}
end{align}$$
We can see that $mathcal{P}$ is a Baire partition. Fix $B_1 cap dots cap B_n in mathcal{P}$. We have $B_1 cap dots cap B_n subseteq G_{x_j}$ for some $G_{x_j}$. If not, we would have $$B_1 cap dots cap B_n = (X setminus G_{x_1}) cap dots cap (X setminus G_{x_n}) = emptyset$$ contradicting $B_1 cap dots cap B_n$ being nonempty. Hence $$begin{align}
sup_{y, z in B_1 cap dots cap B_n} |f(y) - f(z)| &le sup_{y, z in G_{x_j}} |f(y) - f(z)| \
&< varepsilon
end{align}$$
Done!
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
add a comment |
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1 Answer
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Using the hint, our plan is to first find a finite open cover, followed by a finite compact $G_delta$ cover and finally a Baire partition.
Let $varepsilon > 0$ and $x in X$. By continuity of $f$, we can find an open $U_x$ containing $x$ such that for all $u in U_x$ we have
$$|f(x) - f(u)| < frac{varepsilon}{4}$$
This implies for all $y, z in U_x$ we get
$$begin{align}
|f(y) - f(z)| &le |f(y) - f(x)| + |f(x) - f(z)| \
&< frac{varepsilon}{4} + frac{varepsilon}{4} \
&= frac{varepsilon}{2}
end{align}$$
By order preserving property of supremum, we obtain
$$sup_{y, z in U_x} |f(y) - f(z)| le frac{varepsilon}{2} < varepsilon$$
Next, since $X$ is locally compact Hausdorff, we can choose a compact neighbourhood $K_x$ and open $V_x$ such that $$x in V_x subseteq K_x subseteq U_x$$
Now by lemma 3 of this question (please check!),
we can pick a compact $G_delta$ set $G_x$ such that $$x in V_x subseteq K_x subseteq G_x subseteq U_x$$
Observe that ${V_x}_{x in X}$ is an open cover for $X$.
By compactness of $X$, there is a finite subcover ${V_{x_1}, dots, V_{x_n}}$. Since each $V_{x_j} subseteq G_{x_j}$, ${G_{x_1}, dots, G_{x_n}}$ is also a finite cover. Hence ${G_{x_1}, dots, G_{x_n}}$ is a finite compact $G_delta$ cover. Besides, since each $G_{x_j} subseteq U_{x_j}$, we have $$sup_{y, z in G_{x_j}} |f(y) - f(z)| < varepsilon$$
Finally, let $$begin{align}
mathcal{P} = {&B_1 cap dots cap B_n mid \
&B_j = G_{x_j} text{ or } X setminus G_{x_j} text{ with } B_1 cap dots cap B_n neq emptyset}
end{align}$$
We can see that $mathcal{P}$ is a Baire partition. Fix $B_1 cap dots cap B_n in mathcal{P}$. We have $B_1 cap dots cap B_n subseteq G_{x_j}$ for some $G_{x_j}$. If not, we would have $$B_1 cap dots cap B_n = (X setminus G_{x_1}) cap dots cap (X setminus G_{x_n}) = emptyset$$ contradicting $B_1 cap dots cap B_n$ being nonempty. Hence $$begin{align}
sup_{y, z in B_1 cap dots cap B_n} |f(y) - f(z)| &le sup_{y, z in G_{x_j}} |f(y) - f(z)| \
&< varepsilon
end{align}$$
Done!
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
add a comment |
Using the hint, our plan is to first find a finite open cover, followed by a finite compact $G_delta$ cover and finally a Baire partition.
Let $varepsilon > 0$ and $x in X$. By continuity of $f$, we can find an open $U_x$ containing $x$ such that for all $u in U_x$ we have
$$|f(x) - f(u)| < frac{varepsilon}{4}$$
This implies for all $y, z in U_x$ we get
$$begin{align}
|f(y) - f(z)| &le |f(y) - f(x)| + |f(x) - f(z)| \
&< frac{varepsilon}{4} + frac{varepsilon}{4} \
&= frac{varepsilon}{2}
end{align}$$
By order preserving property of supremum, we obtain
$$sup_{y, z in U_x} |f(y) - f(z)| le frac{varepsilon}{2} < varepsilon$$
Next, since $X$ is locally compact Hausdorff, we can choose a compact neighbourhood $K_x$ and open $V_x$ such that $$x in V_x subseteq K_x subseteq U_x$$
Now by lemma 3 of this question (please check!),
we can pick a compact $G_delta$ set $G_x$ such that $$x in V_x subseteq K_x subseteq G_x subseteq U_x$$
Observe that ${V_x}_{x in X}$ is an open cover for $X$.
By compactness of $X$, there is a finite subcover ${V_{x_1}, dots, V_{x_n}}$. Since each $V_{x_j} subseteq G_{x_j}$, ${G_{x_1}, dots, G_{x_n}}$ is also a finite cover. Hence ${G_{x_1}, dots, G_{x_n}}$ is a finite compact $G_delta$ cover. Besides, since each $G_{x_j} subseteq U_{x_j}$, we have $$sup_{y, z in G_{x_j}} |f(y) - f(z)| < varepsilon$$
Finally, let $$begin{align}
mathcal{P} = {&B_1 cap dots cap B_n mid \
&B_j = G_{x_j} text{ or } X setminus G_{x_j} text{ with } B_1 cap dots cap B_n neq emptyset}
end{align}$$
We can see that $mathcal{P}$ is a Baire partition. Fix $B_1 cap dots cap B_n in mathcal{P}$. We have $B_1 cap dots cap B_n subseteq G_{x_j}$ for some $G_{x_j}$. If not, we would have $$B_1 cap dots cap B_n = (X setminus G_{x_1}) cap dots cap (X setminus G_{x_n}) = emptyset$$ contradicting $B_1 cap dots cap B_n$ being nonempty. Hence $$begin{align}
sup_{y, z in B_1 cap dots cap B_n} |f(y) - f(z)| &le sup_{y, z in G_{x_j}} |f(y) - f(z)| \
&< varepsilon
end{align}$$
Done!
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
add a comment |
Using the hint, our plan is to first find a finite open cover, followed by a finite compact $G_delta$ cover and finally a Baire partition.
Let $varepsilon > 0$ and $x in X$. By continuity of $f$, we can find an open $U_x$ containing $x$ such that for all $u in U_x$ we have
$$|f(x) - f(u)| < frac{varepsilon}{4}$$
This implies for all $y, z in U_x$ we get
$$begin{align}
|f(y) - f(z)| &le |f(y) - f(x)| + |f(x) - f(z)| \
&< frac{varepsilon}{4} + frac{varepsilon}{4} \
&= frac{varepsilon}{2}
end{align}$$
By order preserving property of supremum, we obtain
$$sup_{y, z in U_x} |f(y) - f(z)| le frac{varepsilon}{2} < varepsilon$$
Next, since $X$ is locally compact Hausdorff, we can choose a compact neighbourhood $K_x$ and open $V_x$ such that $$x in V_x subseteq K_x subseteq U_x$$
Now by lemma 3 of this question (please check!),
we can pick a compact $G_delta$ set $G_x$ such that $$x in V_x subseteq K_x subseteq G_x subseteq U_x$$
Observe that ${V_x}_{x in X}$ is an open cover for $X$.
By compactness of $X$, there is a finite subcover ${V_{x_1}, dots, V_{x_n}}$. Since each $V_{x_j} subseteq G_{x_j}$, ${G_{x_1}, dots, G_{x_n}}$ is also a finite cover. Hence ${G_{x_1}, dots, G_{x_n}}$ is a finite compact $G_delta$ cover. Besides, since each $G_{x_j} subseteq U_{x_j}$, we have $$sup_{y, z in G_{x_j}} |f(y) - f(z)| < varepsilon$$
Finally, let $$begin{align}
mathcal{P} = {&B_1 cap dots cap B_n mid \
&B_j = G_{x_j} text{ or } X setminus G_{x_j} text{ with } B_1 cap dots cap B_n neq emptyset}
end{align}$$
We can see that $mathcal{P}$ is a Baire partition. Fix $B_1 cap dots cap B_n in mathcal{P}$. We have $B_1 cap dots cap B_n subseteq G_{x_j}$ for some $G_{x_j}$. If not, we would have $$B_1 cap dots cap B_n = (X setminus G_{x_1}) cap dots cap (X setminus G_{x_n}) = emptyset$$ contradicting $B_1 cap dots cap B_n$ being nonempty. Hence $$begin{align}
sup_{y, z in B_1 cap dots cap B_n} |f(y) - f(z)| &le sup_{y, z in G_{x_j}} |f(y) - f(z)| \
&< varepsilon
end{align}$$
Done!
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
Using the hint, our plan is to first find a finite open cover, followed by a finite compact $G_delta$ cover and finally a Baire partition.
Let $varepsilon > 0$ and $x in X$. By continuity of $f$, we can find an open $U_x$ containing $x$ such that for all $u in U_x$ we have
$$|f(x) - f(u)| < frac{varepsilon}{4}$$
This implies for all $y, z in U_x$ we get
$$begin{align}
|f(y) - f(z)| &le |f(y) - f(x)| + |f(x) - f(z)| \
&< frac{varepsilon}{4} + frac{varepsilon}{4} \
&= frac{varepsilon}{2}
end{align}$$
By order preserving property of supremum, we obtain
$$sup_{y, z in U_x} |f(y) - f(z)| le frac{varepsilon}{2} < varepsilon$$
Next, since $X$ is locally compact Hausdorff, we can choose a compact neighbourhood $K_x$ and open $V_x$ such that $$x in V_x subseteq K_x subseteq U_x$$
Now by lemma 3 of this question (please check!),
we can pick a compact $G_delta$ set $G_x$ such that $$x in V_x subseteq K_x subseteq G_x subseteq U_x$$
Observe that ${V_x}_{x in X}$ is an open cover for $X$.
By compactness of $X$, there is a finite subcover ${V_{x_1}, dots, V_{x_n}}$. Since each $V_{x_j} subseteq G_{x_j}$, ${G_{x_1}, dots, G_{x_n}}$ is also a finite cover. Hence ${G_{x_1}, dots, G_{x_n}}$ is a finite compact $G_delta$ cover. Besides, since each $G_{x_j} subseteq U_{x_j}$, we have $$sup_{y, z in G_{x_j}} |f(y) - f(z)| < varepsilon$$
Finally, let $$begin{align}
mathcal{P} = {&B_1 cap dots cap B_n mid \
&B_j = G_{x_j} text{ or } X setminus G_{x_j} text{ with } B_1 cap dots cap B_n neq emptyset}
end{align}$$
We can see that $mathcal{P}$ is a Baire partition. Fix $B_1 cap dots cap B_n in mathcal{P}$. We have $B_1 cap dots cap B_n subseteq G_{x_j}$ for some $G_{x_j}$. If not, we would have $$B_1 cap dots cap B_n = (X setminus G_{x_1}) cap dots cap (X setminus G_{x_n}) = emptyset$$ contradicting $B_1 cap dots cap B_n$ being nonempty. Hence $$begin{align}
sup_{y, z in B_1 cap dots cap B_n} |f(y) - f(z)| &le sup_{y, z in G_{x_j}} |f(y) - f(z)| \
&< varepsilon
end{align}$$
Done!
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
edited Nov 21 '18 at 6:28
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
answered Nov 20 '18 at 23:20
Alex Vong
1,288819
1,288819
add a comment |
add a comment |
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Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 '18 at 0:42
The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 '18 at 0:57
@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 '18 at 1:10
I already put the definitions in the statement.
– eraldcoil
Nov 20 '18 at 1:15