Non-Borel set in arbitrary metric spaceDerived Sets in arbitrary metric space$A subseteq (X,d)$ is compact....
Put the phone down / Put down the phone
What is the tangent at a sharp point on a curve?
Weird lines in Microsoft Word
Is there a POSIX way to shutdown a UNIX machine?
Pre-Employment Background Check With Consent For Future Checks
How can I, as DM, avoid the Conga Line of Death occurring when implementing some form of flanking rule?
Non-Borel set in arbitrary metric space
How do I lift the insulation blower into the attic?
Why doesn't Gödel's incompleteness theorem apply to false statements?
How to get directions in deep space?
Why is implicit conversion not ambiguous for non-primitive types?
Checking @@ROWCOUNT failing
Travelling in US for more than 90 days
Offset in split text content
"Oh no!" in Latin
How do you justify more code being written by following clean code practices?
Extract substring according to regexp with sed or grep
Started in 1987 vs. Starting in 1987
Why does the frost depth increase when the surface temperature warms up?
Magnifying glass in hyperbolic space
Not hide and seek
Highest stage count that are used one right after the other?
Is this saw blade faulty?
Rendered textures different to 3D View
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 4 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 4 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 4 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 4 hours ago
Daniel LiDaniel Li
752414
asked 4 hours ago
Daniel LiDaniel Li
752414
asked 4 hours ago
Daniel LiDaniel Li
752414
asked 4 hours ago
Daniel LiDaniel Li
752414
asked 4 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 3 hours ago
MartinMartin
1,096917
$endgroup$
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 2 hours ago
Noah SchweberNoah Schweber
127k10151290
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154781%2fnon-borel-set-in-arbitrary-metric-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
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 $mathbb{R}$ 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 3 hours ago
MartinMartin
1,096917
$endgroup$
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 3 hours ago
MartinMartin
1,096917
$endgroup$
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 3 hours ago
MartinMartin
1,096917
$endgroup$
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 3 hours ago
MartinMartin
1,096917
edited 3 hours ago
answered 3 hours ago
MartinMartin
1,096917
answered 3 hours ago
MartinMartin
1,096917
answered 3 hours ago
MartinMartin
1,096917
1,096917
add a comment |
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 2 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 2 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 2 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 2 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 2 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 2 hours ago
Noah SchweberNoah Schweber
127k10151290
answered 2 hours ago
Noah SchweberNoah Schweber
127k10151290
127k10151290
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154781%2fnon-borel-set-in-arbitrary-metric-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
