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?
Limit max CPU usage SQL SERVER with WSRM
"Oh no!" in Latin
Can I cause damage to electrical appliances by unplugging them when they are turned on?
Proving an identity involving cross products and coplanar vectors
Can I say "fingers" when referring to toes?
Deciphering cause of death?
Showing mass murder in a kid's book
Unable to disable Microsoft Store in domain environment
Should I warn a new PhD Student?
Is there a distance limit for minecart tracks?
Determining multivariate least squares with constraint
How to preserve electronics (computers, iPads and phones) for hundreds of years
SOQL query causes internal Salesforce error
How to understand "he realized a split second too late was also a mistake"
Given this phrasing in the lease, when should I pay my rent?
ContourPlot — How do I color by contour curvature?
How to leave product feedback on macOS?
Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?
Did I make a mistake by ccing email to boss to others?
Ways of geometrical multiplication
Overlapping circles covering polygon
Make a Bowl of Alphabet Soup
Why the "ls" command is showing the permissions of files in a FAT32 partition?
What the heck is gets(stdin) on site coderbyte?
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 yesterday
Daniel LiDaniel Li
779414
$endgroup$
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 yesterday
Daniel LiDaniel Li
779414
$endgroup$
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 yesterday
Daniel LiDaniel Li
779414
$endgroup$
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 yesterday
Daniel LiDaniel Li
779414
asked yesterday
Daniel LiDaniel Li
779414
asked yesterday
Daniel LiDaniel Li
779414
asked yesterday
Daniel LiDaniel Li
779414
asked yesterday
Daniel LiDaniel Li
779414
779414
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 yesterday
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
21 hours 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 yesterday
Noah SchweberNoah Schweber
127k10151291
$endgroup$
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 $mathbbR$ 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 yesterday
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
21 hours 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 $mathbbR$ 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 yesterday
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
21 hours 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 $mathbbR$ 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 yesterday
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 yesterday
MartinMartin
1,1811019
edited yesterday
answered yesterday
MartinMartin
1,1811019
answered yesterday
MartinMartin
1,1811019
answered yesterday
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
21 hours 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
21 hours 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
21 hours 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
21 hours 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 yesterday
Noah SchweberNoah Schweber
127k10151291
$endgroup$
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 yesterday
Noah SchweberNoah Schweber
127k10151291
$endgroup$
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 yesterday
Noah SchweberNoah Schweber
127k10151291
$endgroup$
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 yesterday
Noah SchweberNoah Schweber
127k10151291
answered yesterday
Noah SchweberNoah Schweber
127k10151291
answered yesterday
Noah SchweberNoah Schweber
127k10151291
answered yesterday
Noah SchweberNoah Schweber
127k10151291
127k10151291
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
