Non-Borel set in arbitrary metric spaceDerived Sets in arbitrary metric space$A subseteq (X,d)$ is compact....
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Non-Borel set in arbitrary metric space
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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 5 hours ago
Daniel LiDaniel Li
752414
$endgroup$
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 5 hours ago
Daniel LiDaniel Li
752414
$endgroup$
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 5 hours ago
Daniel LiDaniel Li
752414
$endgroup$
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 5 hours ago
Daniel LiDaniel Li
752414
asked 5 hours ago
Daniel LiDaniel Li
752414
asked 5 hours ago
Daniel LiDaniel Li
752414
asked 5 hours ago
Daniel LiDaniel Li
752414
asked 5 hours ago
Daniel LiDaniel Li
752414
752414
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 $mathbb{R}$ 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 $d:{x,y}times {x,y} to {0,1}$ 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
$$
{{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
$$
where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.
answered 4 hours ago
MartinMartin
1,106917
$endgroup$
$begingroup$
+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
22 mins ago
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 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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active
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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 $mathbb{R}$ 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 $d:{x,y}times {x,y} to {0,1}$ 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
$$
{{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
$$
where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.
answered 4 hours ago
MartinMartin
1,106917
$endgroup$
$begingroup$
+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
22 mins ago
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 $mathbb{R}$ 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 $d:{x,y}times {x,y} to {0,1}$ 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
$$
{{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
$$
where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.
answered 4 hours ago
MartinMartin
1,106917
$endgroup$
$begingroup$
+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
22 mins ago
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 $mathbb{R}$ 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 $d:{x,y}times {x,y} to {0,1}$ 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
$$
{{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
$$
where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.
answered 4 hours ago
MartinMartin
1,106917
$endgroup$
Yes, there is indeed examples of non-Borel sets in $C[0,1]$ of all continuous functions from $[0,1]$ to $mathbb{R}$ 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 $d:{x,y}times {x,y} to {0,1}$ 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
$$
{{x},{y},{x,y},emptyset} = mathcal{P}({x,y})
$$
where $mathcal{P}({x,y})$ is the powerset of ${x,y}$, so all subsets are Borel measurable sets.
answered 4 hours ago
MartinMartin
1,106917
edited 4 hours ago
answered 4 hours ago
MartinMartin
1,106917
answered 4 hours ago
MartinMartin
1,106917
answered 4 hours ago
MartinMartin
1,106917
1,106917
$begingroup$
+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
22 mins ago
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
22 mins ago
$begingroup$
+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
22 mins ago
$begingroup$
+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
22 mins ago
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 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
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 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
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 3 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
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 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 3 hours ago
Noah SchweberNoah Schweber
127k10151290
127k10151290
add a comment |
add a comment |
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