Problem 2. A comprehensive course in Analysis. Barry simon. 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) $bigcup_{Pinmathcal{P}}P=X$



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.?










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  • Can you define Baire partition and Baire sets for the readers?
    – user25959
    Nov 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 at 0:57










  • @eraldcoil Can you show Hint?
    – Alex Vong
    Nov 20 at 1:10










  • I already put the definitions in the statement.
    – eraldcoil
    Nov 20 at 1:15















up vote
1
down vote

favorite
1












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.?










share|cite|improve this question
























  • Can you define Baire partition and Baire sets for the readers?
    – user25959
    Nov 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 at 0:57










  • @eraldcoil Can you show Hint?
    – Alex Vong
    Nov 20 at 1:10










  • I already put the definitions in the statement.
    – eraldcoil
    Nov 20 at 1:15













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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.?










share|cite|improve this question















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






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edited Nov 18 at 2:22









Nate Eldredge

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asked Nov 18 at 0:38









eraldcoil

26119




26119












  • Can you define Baire partition and Baire sets for the readers?
    – user25959
    Nov 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 at 0:57










  • @eraldcoil Can you show Hint?
    – Alex Vong
    Nov 20 at 1:10










  • I already put the definitions in the statement.
    – eraldcoil
    Nov 20 at 1:15


















  • Can you define Baire partition and Baire sets for the readers?
    – user25959
    Nov 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 at 0:57










  • @eraldcoil Can you show Hint?
    – Alex Vong
    Nov 20 at 1:10










  • I already put the definitions in the statement.
    – eraldcoil
    Nov 20 at 1:15
















Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 18 at 0:42




Can you define Baire partition and Baire sets for the readers?
– user25959
Nov 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 at 0:57




The book does not define it exactly. But I will put the definition that gives partition and Baire.
– eraldcoil
Nov 18 at 0:57












@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 at 1:10




@eraldcoil Can you show Hint?
– Alex Vong
Nov 20 at 1:10












I already put the definitions in the statement.
– eraldcoil
Nov 20 at 1:15




I already put the definitions in the statement.
– eraldcoil
Nov 20 at 1:15










1 Answer
1






active

oldest

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up vote
1
down vote



accepted










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!






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    1 Answer
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    up vote
    1
    down vote



    accepted










    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!






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      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!






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        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!






        share|cite|improve this answer














        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!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 6:28

























        answered Nov 20 at 23:20









        Alex Vong

        1,137819




        1,137819






























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