Trjtdtk

What should you do if you miss a job interview (deliberately)?

Why are synthetic pH indicators used over natural indicators?

Store Credit Card Information in Password Manager?

Should I outline or discovery write my stories?

Can the Supreme Court overturn an impeachment?

Is this toilet slogan correct usage of the English language?

Loading commands from file

why `nmap 192.168.1.97` returns less services than `nmap 127.0.0.1`?

Lowest total scrabble score

Reply 'no position' while the job posting is still there

How can "mimic phobia" be cured or prevented?

Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute?

Melting point of aspirin, contradicting sources

Is the U.S. Code copyrighted by the Government?

Create all possible words using a set or letters

Is it possible to have a strip of cold climate in the middle of a planet?

Count the occurrence of each unique word in the file

By means of an example, show that P(A) + P(B) = 1 does not mean that B is the complement of A.

MTG Artifact and Enchantment Rulings

Freedom of speech and where it applies

How is flyblackbird.com operating under Part 91K?

Why do we read the Megillah by night and by day?

How to explain what's wrong with this application of the chain rule?

Writing bit difficult equation in latex

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?

6

1

$begingroup$

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

share|cite|improve this question

asked Mar 19 at 23:34

Daniel LiDaniel Li

786414

$endgroup$

add a comment | 

6

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 19 at 23:34

Daniel LiDaniel Li

786414

$endgroup$

share|cite|improve this question

asked Mar 19 at 23:34

Daniel LiDaniel Li

786414

share|cite|improve this question

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

2 Answers
2

active

oldest

votes

11

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:27

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 | 

10

$begingroup$

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.

share|cite|improve this answer

answered Mar 20 at 1:12

Noah SchweberNoah Schweber

127k10151292

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

11

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:27

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 | 

11

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:27

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.

share|cite|improve this answer

edited Mar 20 at 0:27

answered Mar 20 at 0:16

MartinMartin

1,1811019

$endgroup$

share|cite|improve this answer

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

10

$begingroup$

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.

share|cite|improve this answer

answered Mar 20 at 1:12

Noah SchweberNoah Schweber

127k10151292

$endgroup$

add a comment | 

10

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 20 at 1:12

Noah SchweberNoah Schweber

127k10151292

$endgroup$

share|cite|improve this answer

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