Are the path components Borel sets?

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It is know that the connected components of a topological space are closed, and that the path connected components (pcc) need not to be closed, what I was wondering is if the pcc are always Borel sets, i.e. the pcc all belong to the $sigma$-algebra generates by the topology.




general-topology path-connected borel-sets




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  • Are you interested in particular topological spaces?
    – Michael Greinecker♦
    Jul 28 at 16:52














up vote
4
down vote

favorite
2












It is know that the connected components of a topological space are closed, and that the path connected components (pcc) need not to be closed, what I was wondering is if the pcc are always Borel sets, i.e. the pcc all belong to the $sigma$-algebra generates by the topology.




general-topology path-connected borel-sets




share|cite|improve this question



















  • Are you interested in particular topological spaces?
    – Michael Greinecker♦
    Jul 28 at 16:52












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





It is know that the connected components of a topological space are closed, and that the path connected components (pcc) need not to be closed, what I was wondering is if the pcc are always Borel sets, i.e. the pcc all belong to the $sigma$-algebra generates by the topology.




general-topology path-connected borel-sets




share|cite|improve this question











It is know that the connected components of a topological space are closed, and that the path connected components (pcc) need not to be closed, what I was wondering is if the pcc are always Borel sets, i.e. the pcc all belong to the $sigma$-algebra generates by the topology.





general-topology path-connected borel-sets





share|cite|improve this question










share|cite|improve this question




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asked Jul 28 at 15:58









G. Ottaviano

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17710











  • Are you interested in particular topological spaces?
    – Michael Greinecker♦
    Jul 28 at 16:52
















  • Are you interested in particular topological spaces?
    – Michael Greinecker♦
    Jul 28 at 16:52















Are you interested in particular topological spaces?
– Michael Greinecker♦
Jul 28 at 16:52




Are you interested in particular topological spaces?
– Michael Greinecker♦
Jul 28 at 16:52










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No. For instance, let $I=mathbbRsetminusmathbbQ$ and let $Asubset I$ be any non-Borel set. Let $BsubsetmathbbR^2$ be union of the line segments connecting $(0,1)$ to each point of $Atimes0$ and the line segments connecting $(0,-1)$ to each point of $(Isetminus A)times0$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $Itimes0$ would be Borel in $Itimes0$. But that intersection is just $Atimes0$, which is not Borel.






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    1 Answer
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    1 Answer
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    active

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    up vote
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    accepted










    No. For instance, let $I=mathbbRsetminusmathbbQ$ and let $Asubset I$ be any non-Borel set. Let $BsubsetmathbbR^2$ be union of the line segments connecting $(0,1)$ to each point of $Atimes0$ and the line segments connecting $(0,-1)$ to each point of $(Isetminus A)times0$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $Itimes0$ would be Borel in $Itimes0$. But that intersection is just $Atimes0$, which is not Borel.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      No. For instance, let $I=mathbbRsetminusmathbbQ$ and let $Asubset I$ be any non-Borel set. Let $BsubsetmathbbR^2$ be union of the line segments connecting $(0,1)$ to each point of $Atimes0$ and the line segments connecting $(0,-1)$ to each point of $(Isetminus A)times0$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $Itimes0$ would be Borel in $Itimes0$. But that intersection is just $Atimes0$, which is not Borel.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        No. For instance, let $I=mathbbRsetminusmathbbQ$ and let $Asubset I$ be any non-Borel set. Let $BsubsetmathbbR^2$ be union of the line segments connecting $(0,1)$ to each point of $Atimes0$ and the line segments connecting $(0,-1)$ to each point of $(Isetminus A)times0$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $Itimes0$ would be Borel in $Itimes0$. But that intersection is just $Atimes0$, which is not Borel.






        share|cite|improve this answer













        No. For instance, let $I=mathbbRsetminusmathbbQ$ and let $Asubset I$ be any non-Borel set. Let $BsubsetmathbbR^2$ be union of the line segments connecting $(0,1)$ to each point of $Atimes0$ and the line segments connecting $(0,-1)$ to each point of $(Isetminus A)times0$. Then $B$ has two path components, the union $C$ of all the line segments from $(0,1)$ and the union $D$ of all the line segments from $(0,-1)$. If $C$ were Borel in $B$, then its intersection with $Itimes0$ would be Borel in $Itimes0$. But that intersection is just $Atimes0$, which is not Borel.







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        answered Jul 29 at 0:41









        Eric Wofsey

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