Compact spaces in which any closed set can be partitioned into finitely many closed sets whose clopen subsets...












6












$begingroup$


Let $X $ be a compact topological space (not necessarilly Hausdorff). I am looking for a charactrization for the following property:



Property: If $C $ is a closed subset of $X $, then there are pairwise disjoint closed subsets $C_1$,...,$C_n $ of $X $ such that $C=C_1cupdotscup C_n $, and each $C_i $ has the property that if $A $ is a clopen subest of $C_i $, then there exists a clopen subset $B $ of $X $ with $A=C_icap B$?



Any comment is very welcome.










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$endgroup$








  • 4




    $begingroup$
    The spaces that every closed subset has finitely many connected components are example of such spaces.
    $endgroup$
    – Es.Ro
    Jan 16 at 17:21






  • 2




    $begingroup$
    Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
    $endgroup$
    – Eric Wofsey
    Jan 16 at 19:02










  • $begingroup$
    By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
    $endgroup$
    – Luke
    Jan 16 at 22:29










  • $begingroup$
    @Luke, yes in the subspace topology.
    $endgroup$
    – Andy.G
    Jan 17 at 2:35






  • 1




    $begingroup$
    Neither direction is obvious; I'm writing an answer elaborating now.
    $endgroup$
    – Eric Wofsey
    Jan 17 at 2:51
















6












$begingroup$


Let $X $ be a compact topological space (not necessarilly Hausdorff). I am looking for a charactrization for the following property:



Property: If $C $ is a closed subset of $X $, then there are pairwise disjoint closed subsets $C_1$,...,$C_n $ of $X $ such that $C=C_1cupdotscup C_n $, and each $C_i $ has the property that if $A $ is a clopen subest of $C_i $, then there exists a clopen subset $B $ of $X $ with $A=C_icap B$?



Any comment is very welcome.










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    The spaces that every closed subset has finitely many connected components are example of such spaces.
    $endgroup$
    – Es.Ro
    Jan 16 at 17:21






  • 2




    $begingroup$
    Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
    $endgroup$
    – Eric Wofsey
    Jan 16 at 19:02










  • $begingroup$
    By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
    $endgroup$
    – Luke
    Jan 16 at 22:29










  • $begingroup$
    @Luke, yes in the subspace topology.
    $endgroup$
    – Andy.G
    Jan 17 at 2:35






  • 1




    $begingroup$
    Neither direction is obvious; I'm writing an answer elaborating now.
    $endgroup$
    – Eric Wofsey
    Jan 17 at 2:51














6












6








6


2



$begingroup$


Let $X $ be a compact topological space (not necessarilly Hausdorff). I am looking for a charactrization for the following property:



Property: If $C $ is a closed subset of $X $, then there are pairwise disjoint closed subsets $C_1$,...,$C_n $ of $X $ such that $C=C_1cupdotscup C_n $, and each $C_i $ has the property that if $A $ is a clopen subest of $C_i $, then there exists a clopen subset $B $ of $X $ with $A=C_icap B$?



Any comment is very welcome.










share|cite|improve this question











$endgroup$




Let $X $ be a compact topological space (not necessarilly Hausdorff). I am looking for a charactrization for the following property:



Property: If $C $ is a closed subset of $X $, then there are pairwise disjoint closed subsets $C_1$,...,$C_n $ of $X $ such that $C=C_1cupdotscup C_n $, and each $C_i $ has the property that if $A $ is a clopen subest of $C_i $, then there exists a clopen subset $B $ of $X $ with $A=C_icap B$?



Any comment is very welcome.







general-topology compactness connectedness






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 2:59









Eric Wofsey

192k14220352




192k14220352










asked Jan 16 at 17:01









Andy.GAndy.G

935




935








  • 4




    $begingroup$
    The spaces that every closed subset has finitely many connected components are example of such spaces.
    $endgroup$
    – Es.Ro
    Jan 16 at 17:21






  • 2




    $begingroup$
    Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
    $endgroup$
    – Eric Wofsey
    Jan 16 at 19:02










  • $begingroup$
    By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
    $endgroup$
    – Luke
    Jan 16 at 22:29










  • $begingroup$
    @Luke, yes in the subspace topology.
    $endgroup$
    – Andy.G
    Jan 17 at 2:35






  • 1




    $begingroup$
    Neither direction is obvious; I'm writing an answer elaborating now.
    $endgroup$
    – Eric Wofsey
    Jan 17 at 2:51














  • 4




    $begingroup$
    The spaces that every closed subset has finitely many connected components are example of such spaces.
    $endgroup$
    – Es.Ro
    Jan 16 at 17:21






  • 2




    $begingroup$
    Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
    $endgroup$
    – Eric Wofsey
    Jan 16 at 19:02










  • $begingroup$
    By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
    $endgroup$
    – Luke
    Jan 16 at 22:29










  • $begingroup$
    @Luke, yes in the subspace topology.
    $endgroup$
    – Andy.G
    Jan 17 at 2:35






  • 1




    $begingroup$
    Neither direction is obvious; I'm writing an answer elaborating now.
    $endgroup$
    – Eric Wofsey
    Jan 17 at 2:51








4




4




$begingroup$
The spaces that every closed subset has finitely many connected components are example of such spaces.
$endgroup$
– Es.Ro
Jan 16 at 17:21




$begingroup$
The spaces that every closed subset has finitely many connected components are example of such spaces.
$endgroup$
– Es.Ro
Jan 16 at 17:21




2




2




$begingroup$
Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
$endgroup$
– Eric Wofsey
Jan 16 at 19:02




$begingroup$
Well, in the case that $X$ is Hausdorff, this is equivalent to total disconnectedness.
$endgroup$
– Eric Wofsey
Jan 16 at 19:02












$begingroup$
By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
$endgroup$
– Luke
Jan 16 at 22:29




$begingroup$
By “clopen subset of $C_i$”, you surely mean “subset clopen w.r.t. the subspace topology of $C_i$”?
$endgroup$
– Luke
Jan 16 at 22:29












$begingroup$
@Luke, yes in the subspace topology.
$endgroup$
– Andy.G
Jan 17 at 2:35




$begingroup$
@Luke, yes in the subspace topology.
$endgroup$
– Andy.G
Jan 17 at 2:35




1




1




$begingroup$
Neither direction is obvious; I'm writing an answer elaborating now.
$endgroup$
– Eric Wofsey
Jan 17 at 2:51




$begingroup$
Neither direction is obvious; I'm writing an answer elaborating now.
$endgroup$
– Eric Wofsey
Jan 17 at 2:51










1 Answer
1






active

oldest

votes


















3












$begingroup$

I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.



To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $Csubseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $Csetminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.



Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.



To prove this, pick an infinite discrete subset $Dsubset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=overline{D}$. Suppose a decomposition $C=C_1cupdotscup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $Asubset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $Bcap C_i$ for any clopen $Bsubseteq X$.





Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer.
    $endgroup$
    – Andy.G
    Jan 17 at 3:00










  • $begingroup$
    Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
    $endgroup$
    – Andy.G
    Jan 17 at 14:55














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.



To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $Csubseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $Csetminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.



Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.



To prove this, pick an infinite discrete subset $Dsubset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=overline{D}$. Suppose a decomposition $C=C_1cupdotscup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $Asubset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $Bcap C_i$ for any clopen $Bsubseteq X$.





Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer.
    $endgroup$
    – Andy.G
    Jan 17 at 3:00










  • $begingroup$
    Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
    $endgroup$
    – Andy.G
    Jan 17 at 14:55


















3












$begingroup$

I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.



To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $Csubseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $Csetminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.



Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.



To prove this, pick an infinite discrete subset $Dsubset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=overline{D}$. Suppose a decomposition $C=C_1cupdotscup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $Asubset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $Bcap C_i$ for any clopen $Bsubseteq X$.





Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer.
    $endgroup$
    – Andy.G
    Jan 17 at 3:00










  • $begingroup$
    Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
    $endgroup$
    – Andy.G
    Jan 17 at 14:55
















3












3








3





$begingroup$

I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.



To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $Csubseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $Csetminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.



Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.



To prove this, pick an infinite discrete subset $Dsubset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=overline{D}$. Suppose a decomposition $C=C_1cupdotscup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $Asubset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $Bcap C_i$ for any clopen $Bsubseteq X$.





Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.






share|cite|improve this answer









$endgroup$



I don't see any reason to believe there is a simpler necessary and sufficient condition in general. When $X$ is Hausdorff there is one, though: your condition is equivalent to $X$ being totally disconnected.



To prove this, first suppose $X$ is compact Hausdorff and totally disconnected. Then clopen sets separate points of $X$ (see Any two points in a Stone space can be disconnected by clopen sets). It follows by compactness that clopen sets also separate closed sets (the proof is the same as the proof that a compact Hausdorff space is normal, just using clopen sets instead of open sets everywhere). So in particular, if $Csubseteq X$ is closed and $A$ is clopen in $C$, then $A$ and $Csetminus A$ can be separated by clopen subsets of $X$ (since they are disjoint and closed in $X$). This means that your condition holds, with $n=1$.



Conversely, suppose $X$ is compact Hausdorff and not totally disconnected. Your property is inherited by closed subspaces, so we may replace $X$ with one of its nontrivial connected components. So, we assume $X$ is compact Hausdorff and connected and has more than one point, and must show it does not satisfy your condition.



To prove this, pick an infinite discrete subset $Dsubset X$ (see Every infinite Hausdorff space has an infinite discrete subspace) and consider $C=overline{D}$. Suppose a decomposition $C=C_1cupdotscup C_n$ with your property existed. Note that each point of $D$ is isolated in $C$, and some $C_i$ must contain infinitely many points of $D$. In particular, some $C_i$ must be disconnected, so there is a nontrivial clopen subset $Asubset C_i$. But $X$ is connected, so it has no nontrivial clopen subsets, and so $A$ cannot be $Bcap C_i$ for any clopen $Bsubseteq X$.





Of course, total disconnectedness is not necessary in the non-Hausdorff case (I don't know whether it is sufficient). Besides trivial examples like the indiscrete topology, there is also the cofinite topology on any set. More generally, as Es.Ro commented, a sufficient condition is that every closed subset of $X$ has only finitely many connected components, since then you can take the connected components as the $C_i$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 2:54









Eric WofseyEric Wofsey

192k14220352




192k14220352












  • $begingroup$
    Thanks for your answer.
    $endgroup$
    – Andy.G
    Jan 17 at 3:00










  • $begingroup$
    Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
    $endgroup$
    – Andy.G
    Jan 17 at 14:55




















  • $begingroup$
    Thanks for your answer.
    $endgroup$
    – Andy.G
    Jan 17 at 3:00










  • $begingroup$
    Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
    $endgroup$
    – Andy.G
    Jan 17 at 14:55


















$begingroup$
Thanks for your answer.
$endgroup$
– Andy.G
Jan 17 at 3:00




$begingroup$
Thanks for your answer.
$endgroup$
– Andy.G
Jan 17 at 3:00












$begingroup$
Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
$endgroup$
– Andy.G
Jan 17 at 14:55






$begingroup$
Can we find a special family $F $ of closed subsets of $X $ (in non Hausdorff case) such that if each element of $F $ has a partition with the stated property, then all close subsets of $X $ have a partition with the stated property?
$endgroup$
– Andy.G
Jan 17 at 14:55




















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