Borel Hierarchy












4












$begingroup$


i'm in trouble with an exercise on Kechris, Classical Descriptive Set Theory. The Theorem 22.4 shows $Sigma_xi^0(X)neqPi_xi^0(X)$ for each ordinal $xilneqomega_1$ and uncountable polish space $X$, using the existence of universal sets. The following exercise is: Show that if X is an uncountable polish space and $lambda$ is a limit ordinal, then: $bigcup_{xilneqlambda}Sigma_xi^0(X)subsetneqDelta_lambda^0(X)$.



The inclusion is obvious. For the inequality, i would like to show that the set $A=bigcup_{ninomega}A_n$, with $A_n$ taken in $Sigma_{xi_n}^0(X)backslashPi_{xi_n}^0(X)$ and $xi_nlneqxi_{n+1}lneqdotslneqlambda$, is in $Delta_lambda^0(X)$ (clearly) but not in the first set. Is this a successfully way? Otherwise, what's the way?






general-topology set-theory descriptive-set-theory borel-sets







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
    $endgroup$
    – Alex Kruckman
    Jan 23 at 0:44












  • $begingroup$
    @NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
    $endgroup$
    – Alex Kruckman
    Jan 23 at 3:56










  • $begingroup$
    @AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 4:07










  • $begingroup$
    The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
    $endgroup$
    – Ajeje
    Jan 23 at 8:40












  • $begingroup$
    @Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 14:45
















4












$begingroup$


i'm in trouble with an exercise on Kechris, Classical Descriptive Set Theory. The Theorem 22.4 shows $Sigma_xi^0(X)neqPi_xi^0(X)$ for each ordinal $xilneqomega_1$ and uncountable polish space $X$, using the existence of universal sets. The following exercise is: Show that if X is an uncountable polish space and $lambda$ is a limit ordinal, then: $bigcup_{xilneqlambda}Sigma_xi^0(X)subsetneqDelta_lambda^0(X)$.



The inclusion is obvious. For the inequality, i would like to show that the set $A=bigcup_{ninomega}A_n$, with $A_n$ taken in $Sigma_{xi_n}^0(X)backslashPi_{xi_n}^0(X)$ and $xi_nlneqxi_{n+1}lneqdotslneqlambda$, is in $Delta_lambda^0(X)$ (clearly) but not in the first set. Is this a successfully way? Otherwise, what's the way?






general-topology set-theory descriptive-set-theory borel-sets







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
    $endgroup$
    – Alex Kruckman
    Jan 23 at 0:44












  • $begingroup$
    @NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
    $endgroup$
    – Alex Kruckman
    Jan 23 at 3:56










  • $begingroup$
    @AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 4:07










  • $begingroup$
    The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
    $endgroup$
    – Ajeje
    Jan 23 at 8:40












  • $begingroup$
    @Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 14:45














4












4








4





$begingroup$


i'm in trouble with an exercise on Kechris, Classical Descriptive Set Theory. The Theorem 22.4 shows $Sigma_xi^0(X)neqPi_xi^0(X)$ for each ordinal $xilneqomega_1$ and uncountable polish space $X$, using the existence of universal sets. The following exercise is: Show that if X is an uncountable polish space and $lambda$ is a limit ordinal, then: $bigcup_{xilneqlambda}Sigma_xi^0(X)subsetneqDelta_lambda^0(X)$.



The inclusion is obvious. For the inequality, i would like to show that the set $A=bigcup_{ninomega}A_n$, with $A_n$ taken in $Sigma_{xi_n}^0(X)backslashPi_{xi_n}^0(X)$ and $xi_nlneqxi_{n+1}lneqdotslneqlambda$, is in $Delta_lambda^0(X)$ (clearly) but not in the first set. Is this a successfully way? Otherwise, what's the way?






general-topology set-theory descriptive-set-theory borel-sets







share|cite|improve this question











$endgroup$




i'm in trouble with an exercise on Kechris, Classical Descriptive Set Theory. The Theorem 22.4 shows $Sigma_xi^0(X)neqPi_xi^0(X)$ for each ordinal $xilneqomega_1$ and uncountable polish space $X$, using the existence of universal sets. The following exercise is: Show that if X is an uncountable polish space and $lambda$ is a limit ordinal, then: $bigcup_{xilneqlambda}Sigma_xi^0(X)subsetneqDelta_lambda^0(X)$.



The inclusion is obvious. For the inequality, i would like to show that the set $A=bigcup_{ninomega}A_n$, with $A_n$ taken in $Sigma_{xi_n}^0(X)backslashPi_{xi_n}^0(X)$ and $xi_nlneqxi_{n+1}lneqdotslneqlambda$, is in $Delta_lambda^0(X)$ (clearly) but not in the first set. Is this a successfully way? Otherwise, what's the way?






general-topology set-theory descriptive-set-theory borel-sets




general-topology set-theory descriptive-set-theory borel-sets






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 5:05









Nate Eldredge

64k682174




64k682174










asked Jan 23 at 0:07









AjejeAjeje

415




415












  • $begingroup$
    I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
    $endgroup$
    – Alex Kruckman
    Jan 23 at 0:44












  • $begingroup$
    @NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
    $endgroup$
    – Alex Kruckman
    Jan 23 at 3:56










  • $begingroup$
    @AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 4:07










  • $begingroup$
    The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
    $endgroup$
    – Ajeje
    Jan 23 at 8:40












  • $begingroup$
    @Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 14:45


















  • $begingroup$
    I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
    $endgroup$
    – Alex Kruckman
    Jan 23 at 0:44












  • $begingroup$
    @NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
    $endgroup$
    – Alex Kruckman
    Jan 23 at 3:56










  • $begingroup$
    @AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 4:07










  • $begingroup$
    The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
    $endgroup$
    – Ajeje
    Jan 23 at 8:40












  • $begingroup$
    @Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
    $endgroup$
    – Nate Eldredge
    Jan 23 at 14:45
















$begingroup$
I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
$endgroup$
– Alex Kruckman
Jan 23 at 0:44






$begingroup$
I guess you mean to take the sequence $(xi_n)_{nin omega}$ to be cofinal in $lambda$? Then $A$ is in $Sigma^0_lambda$, but why should it be in $Delta^0_lambda$? That is, why is it in $Pi^0_lambda$? (You say this is clear, but I don't see it.)
$endgroup$
– Alex Kruckman
Jan 23 at 0:44














$begingroup$
@NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
$endgroup$
– Alex Kruckman
Jan 23 at 3:56




$begingroup$
@NateEldredge Hmm... maybe I'm being dense here, but it seems like that observation only shows that $A$ is a union of sets in various $Pi^0_xi$ classes, so it's in $Sigma^0_lambda$. I'm asking why it's also in $Pi^0_lambda$.
$endgroup$
– Alex Kruckman
Jan 23 at 3:56












$begingroup$
@AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
$endgroup$
– Nate Eldredge
Jan 23 at 4:07




$begingroup$
@AlexKruckman: Nope, it's me who's being dense. Thinking more carefully, I don't see how to prove it either. Sorry.
$endgroup$
– Nate Eldredge
Jan 23 at 4:07












$begingroup$
The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
$endgroup$
– Ajeje
Jan 23 at 8:40






$begingroup$
The $xi_n$ are cofinal in $lambda$, so $A_ninSigma_{xi_n}^0subseteqPi_{xi_{n+1}}^0$ and $xi_{n+1}<lambda$. This was the idea for $AinDelta_lambda^0$. I'm wrong?
$endgroup$
– Ajeje
Jan 23 at 8:40














$begingroup$
@Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
$endgroup$
– Nate Eldredge
Jan 23 at 14:45




$begingroup$
@Ajeje: Yes, and $A$ is the countable union of the $A_n$, so that shows $A in mathbfSigma_lambda^0$. To get it to be in $mathbfDelta_lambda^0$, you have to also show it's in $mathbfPi_lambda^0$, and that means showing that $A^c$ can be written as a countable union of sets $B_n in mathbfSigma_{xi_n}^0$. I don't see how you can show that.
$endgroup$
– Nate Eldredge
Jan 23 at 14:45










1 Answer
1






active

oldest

votes


















0












$begingroup$

Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $bigcup_{xi < lambda} mathbfSigma_xi^0(X)$.



Also, I am unclear how you prove that $A in mathbfPi_lambda^0(X)$ with your construction.



Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n in mathbfPi_{xi_n}^0(U_n) setminus mathbfSigma_{xi_n}^0(U_n)$. Then letting $A = bigcup_n A_n$ as before, it is not too hard to show that $A in mathbf{Delta}_lambda^0(X)$. (Hint: $A^c = (bigcup_n U_n)^c cup bigcup_n (A_n^c cap U_n)$). However, if $A in mathbf{Sigma}_{xi_n}^0(X)$ for some $n$, then $A_n = A cap U_n in mathbfSigma_{xi_n}^0(U_n)$ which is a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
    $endgroup$
    – Ajeje
    Jan 24 at 23:21










  • $begingroup$
    This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:33










  • $begingroup$
    @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:36













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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $bigcup_{xi < lambda} mathbfSigma_xi^0(X)$.



Also, I am unclear how you prove that $A in mathbfPi_lambda^0(X)$ with your construction.



Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n in mathbfPi_{xi_n}^0(U_n) setminus mathbfSigma_{xi_n}^0(U_n)$. Then letting $A = bigcup_n A_n$ as before, it is not too hard to show that $A in mathbf{Delta}_lambda^0(X)$. (Hint: $A^c = (bigcup_n U_n)^c cup bigcup_n (A_n^c cap U_n)$). However, if $A in mathbf{Sigma}_{xi_n}^0(X)$ for some $n$, then $A_n = A cap U_n in mathbfSigma_{xi_n}^0(U_n)$ which is a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
    $endgroup$
    – Ajeje
    Jan 24 at 23:21










  • $begingroup$
    This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:33










  • $begingroup$
    @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:36


















0












$begingroup$

Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $bigcup_{xi < lambda} mathbfSigma_xi^0(X)$.



Also, I am unclear how you prove that $A in mathbfPi_lambda^0(X)$ with your construction.



Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n in mathbfPi_{xi_n}^0(U_n) setminus mathbfSigma_{xi_n}^0(U_n)$. Then letting $A = bigcup_n A_n$ as before, it is not too hard to show that $A in mathbf{Delta}_lambda^0(X)$. (Hint: $A^c = (bigcup_n U_n)^c cup bigcup_n (A_n^c cap U_n)$). However, if $A in mathbf{Sigma}_{xi_n}^0(X)$ for some $n$, then $A_n = A cap U_n in mathbfSigma_{xi_n}^0(U_n)$ which is a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
    $endgroup$
    – Ajeje
    Jan 24 at 23:21










  • $begingroup$
    This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:33










  • $begingroup$
    @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:36
















0












0








0





$begingroup$

Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $bigcup_{xi < lambda} mathbfSigma_xi^0(X)$.



Also, I am unclear how you prove that $A in mathbfPi_lambda^0(X)$ with your construction.



Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n in mathbfPi_{xi_n}^0(U_n) setminus mathbfSigma_{xi_n}^0(U_n)$. Then letting $A = bigcup_n A_n$ as before, it is not too hard to show that $A in mathbf{Delta}_lambda^0(X)$. (Hint: $A^c = (bigcup_n U_n)^c cup bigcup_n (A_n^c cap U_n)$). However, if $A in mathbf{Sigma}_{xi_n}^0(X)$ for some $n$, then $A_n = A cap U_n in mathbfSigma_{xi_n}^0(U_n)$ which is a contradiction.






share|cite|improve this answer











$endgroup$



Your idea doesn't quite work as is; you will have to be more careful in choosing the $A_n$. For instance, as written, it could happen that $xi_1 = 1$ and $A_1$ is an open ball, and all the remaining $A_n$ are contained inside $A_1$. Then $A$ is just $A_1$, which is certainly in $bigcup_{xi < lambda} mathbfSigma_xi^0(X)$.



Also, I am unclear how you prove that $A in mathbfPi_lambda^0(X)$ with your construction.



Here is a modification that will make it easier: "separate" the sets $A_n$. That is, choose a countable family $U_n$ of disjoint uncountable open subsets of $X$ (exercise: show that such a family exists), and let $A_n in mathbfPi_{xi_n}^0(U_n) setminus mathbfSigma_{xi_n}^0(U_n)$. Then letting $A = bigcup_n A_n$ as before, it is not too hard to show that $A in mathbf{Delta}_lambda^0(X)$. (Hint: $A^c = (bigcup_n U_n)^c cup bigcup_n (A_n^c cap U_n)$). However, if $A in mathbf{Sigma}_{xi_n}^0(X)$ for some $n$, then $A_n = A cap U_n in mathbfSigma_{xi_n}^0(U_n)$ which is a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 23 at 14:51

























answered Jan 23 at 4:59









Nate EldredgeNate Eldredge

64k682174




64k682174












  • $begingroup$
    Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
    $endgroup$
    – Ajeje
    Jan 24 at 23:21










  • $begingroup$
    This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:33










  • $begingroup$
    @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:36




















  • $begingroup$
    Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
    $endgroup$
    – Ajeje
    Jan 24 at 23:21










  • $begingroup$
    This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:33










  • $begingroup$
    @Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
    $endgroup$
    – Nate Eldredge
    Jan 24 at 23:36


















$begingroup$
Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
$endgroup$
– Ajeje
Jan 24 at 23:21




$begingroup$
Yes, my idea doesn't work, i was thinking of something similar to your solution. Is there an easier way to solve the problem? When I first read it, i thought it was an easy consequence of the previous theorem
$endgroup$
– Ajeje
Jan 24 at 23:21












$begingroup$
This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
$endgroup$
– Nate Eldredge
Jan 24 at 23:33




$begingroup$
This seems relatively easy to me, and I don't see anything particularly easier. You need to construct a set $A$ that has at least as much "complexity" as each of the sets that make it up. If they can interact with each other, it can all fall apart, and this sort of separation seems to me like the most natural way to prevent that.
$endgroup$
– Nate Eldredge
Jan 24 at 23:33












$begingroup$
@Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
$endgroup$
– Nate Eldredge
Jan 24 at 23:36






$begingroup$
@Ajeje: If you want to be a little more concrete, you can assume without loss of generality that $X$ is Cantor space, and it is clear that $X$ contains countably many disjoint (clopen) copies of Cantor space.
$endgroup$
– Nate Eldredge
Jan 24 at 23:36




















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