Non-Borel set in arbitrary metric spaceDerived Sets in arbitrary metric space$A subseteq (X,d)$ is compact. Which metric $p$ makes $(A times A,p)$ also compact and $d: (A times A,p) rightarrow [0,infty)$ continuous?Borel sets and measurabilityapproximate a Borel set by a continuousAn example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example.A Borel subset of a topological spaceseparability of a metric spacetotally disconnected and non Borel set.What do metric spaces look like?How do we get the notion “Borel regular” measures?
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Non-Borel set in arbitrary metric space
Derived Sets in arbitrary metric space$A subseteq (X,d)$ is compact. Which metric $p$ makes $(A times A,p)$ also compact and $d: (A times A,p) rightarrow [0,infty)$ continuous?Borel sets and measurabilityapproximate a Borel set by a continuousAn example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example.A Borel subset of a topological spaceseparability of a metric spacetotally disconnected and non Borel set.What do metric spaces look like?How do we get the notion “Borel regular” measures?
$begingroup$
Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.
real-analysis general-topology functional-analysis measure-theory
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
$endgroup$
add a comment |
$begingroup$
Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.
real-analysis general-topology functional-analysis measure-theory
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
$endgroup$
add a comment |
$begingroup$
Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.
real-analysis general-topology functional-analysis measure-theory
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
$endgroup$
Most sources give non-Borel set in Euclidean space. I wonder if there is a way to construct such sets in arbitrary metric space. In particular, is there a non-borel set in $C[0,1]$ all continuous functions on $[0,1]$ where metrics is supremum.
real-analysis general-topology functional-analysis measure-theory
real-analysis general-topology functional-analysis measure-theory
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
asked Mar 19 at 23:34
Daniel LiDaniel Li
786414
786414
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbbR$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(x,y,d)$ equipped with the discrete metric $d:x,ytimes x,y to 0,1$ given by
$$
d(x,y)=1, quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
x,y,x,y,emptyset = mathcalP(x,y)
$$
where $mathcalP(x,y)$ is the powerset of $x,y$, so all subsets are Borel measurable sets.
answered Mar 20 at 0:16
MartinMartin
1,1811019
$endgroup$
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
add a comment |
$begingroup$
Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbbR$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(x,y,d)$ equipped with the discrete metric $d:x,ytimes x,y to 0,1$ given by
$$
d(x,y)=1, quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
x,y,x,y,emptyset = mathcalP(x,y)
$$
where $mathcalP(x,y)$ is the powerset of $x,y$, so all subsets are Borel measurable sets.
answered Mar 20 at 0:16
MartinMartin
1,1811019
$endgroup$
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
add a comment |
$begingroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbbR$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(x,y,d)$ equipped with the discrete metric $d:x,ytimes x,y to 0,1$ given by
$$
d(x,y)=1, quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
x,y,x,y,emptyset = mathcalP(x,y)
$$
where $mathcalP(x,y)$ is the powerset of $x,y$, so all subsets are Borel measurable sets.
answered Mar 20 at 0:16
MartinMartin
1,1811019
$endgroup$
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
add a comment |
$begingroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbbR$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(x,y,d)$ equipped with the discrete metric $d:x,ytimes x,y to 0,1$ given by
$$
d(x,y)=1, quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
x,y,x,y,emptyset = mathcalP(x,y)
$$
where $mathcalP(x,y)$ is the powerset of $x,y$, so all subsets are Borel measurable sets.
answered Mar 20 at 0:16
MartinMartin
1,1811019
$endgroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbbR$ equipped with the uniform norm. Namely, the subset of all continuous nowhere differentiable functions is not a Borel set.
This result can be found in:
Mauldin, R. Daniel. The set of continuous nowhere differentiable functions. Pacific J. Math. 83 (1979), no. 1, 199--205.
In regards to the question on whether it is possible to construct non-Borel sets in arbitrary metric spaces, then the answer is no. Consider the metric space $(x,y,d)$ equipped with the discrete metric $d:x,ytimes x,y to 0,1$ given by
$$
d(x,y)=1, quad d(x,x)=d(y,y)=0.
$$
The Borel sigma algebra on this metric space is given by
$$
x,y,x,y,emptyset = mathcalP(x,y)
$$
where $mathcalP(x,y)$ is the powerset of $x,y$, so all subsets are Borel measurable sets.
answered Mar 20 at 0:16
MartinMartin
1,1811019
edited Mar 20 at 0:27
answered Mar 20 at 0:16
MartinMartin
1,1811019
answered Mar 20 at 0:16
MartinMartin
1,1811019
answered Mar 20 at 0:16
MartinMartin
1,1811019
1,1811019
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
add a comment |
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
$begingroup$
+1.... With the discrete metric on any set, all subsets are open, and a fortiori, are Borel. Another example would be any countable metric space $X,$ as any $Ysubset X$ is equal to $ cup y:yin Y,$ which is a countable union of closed sets
$endgroup$
– DanielWainfleet
Mar 20 at 4:18
add a comment |
$begingroup$
Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
$endgroup$
add a comment |
$begingroup$
Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
$endgroup$
add a comment |
$begingroup$
Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
$endgroup$
Martin gave a specific example in $C[0,1]$ and showed that the general example is negative. Let me argue that a broad class of spaces has a positive answer:
In any second-countable topological space, there are only continuum-many Borel sets. Since space with at least continuum many points has more than continuum many subsets, this means that every second-countable space with continuum many points has non-Borel subsets.
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
answered Mar 20 at 1:12
Noah SchweberNoah Schweber
127k10151292
127k10151292
add a comment |
add a comment |
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