if $B in mathscrB(mathbbR^n)$ then $x + B in mathscrB(mathbbR^n) ; forall x in mathbbR^n$

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

I am trying to show that if $B in mathscrB(mathbbR^n)$ then $x + B in mathscrB(mathbbR^n) ; forall x in mathbbR^n$.

To show this, it suffices to show that the set
$$mathscrA_x := Bin mathscrB(mathbbR^n): x + B in mathscrB(mathbbR^n) subset mathscrB(mathbbR^n).$$ I need to show first that $mathscrA_x$ is a $sigma-$algebra. How can I show that if $B in mathscrA_x$ then $B^c in mathscrA_x$, i.e. $x+A^c in mathscrB(mathbbR^n)$?

real-analysis analysis measure-theory

share|cite|improve this question

asked Jul 15 at 5:59

takecare

2,25811431

add a comment | 

up vote
2
down vote

favorite

I am trying to show that if $B in mathscrB(mathbbR^n)$ then $x + B in mathscrB(mathbbR^n) ; forall x in mathbbR^n$.

To show this, it suffices to show that the set
$$mathscrA_x := Bin mathscrB(mathbbR^n): x + B in mathscrB(mathbbR^n) subset mathscrB(mathbbR^n).$$ I need to show first that $mathscrA_x$ is a $sigma-$algebra. How can I show that if $B in mathscrA_x$ then $B^c in mathscrA_x$, i.e. $x+A^c in mathscrB(mathbbR^n)$?

real-analysis analysis measure-theory

share|cite|improve this question

asked Jul 15 at 5:59

takecare

2,25811431

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I am trying to show that if $B in mathscrB(mathbbR^n)$ then $x + B in mathscrB(mathbbR^n) ; forall x in mathbbR^n$.

To show this, it suffices to show that the set
$$mathscrA_x := Bin mathscrB(mathbbR^n): x + B in mathscrB(mathbbR^n) subset mathscrB(mathbbR^n).$$ I need to show first that $mathscrA_x$ is a $sigma-$algebra. How can I show that if $B in mathscrA_x$ then $B^c in mathscrA_x$, i.e. $x+A^c in mathscrB(mathbbR^n)$?

real-analysis analysis measure-theory

share|cite|improve this question

asked Jul 15 at 5:59

takecare

2,25811431

I am trying to show that if $B in mathscrB(mathbbR^n)$ then $x + B in mathscrB(mathbbR^n) ; forall x in mathbbR^n$.

To show this, it suffices to show that the set
$$mathscrA_x := Bin mathscrB(mathbbR^n): x + B in mathscrB(mathbbR^n) subset mathscrB(mathbbR^n).$$ I need to show first that $mathscrA_x$ is a $sigma-$algebra. How can I show that if $B in mathscrA_x$ then $B^c in mathscrA_x$, i.e. $x+A^c in mathscrB(mathbbR^n)$?

real-analysis analysis measure-theory

share|cite|improve this question

asked Jul 15 at 5:59

takecare

2,25811431

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 5:59

takecare

2,25811431

asked Jul 15 at 5:59

takecare

2,25811431

asked Jul 15 at 5:59

takecare

2,25811431

2,25811431

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote



accepted

If $Binmathscr A_x$ then $B=x+A$ for some $Ainmathscr B(mathbb R^n)$.

Then: $$-x+B^complement=(-x+B)^complement=A^complementinmathscr B(mathbb R^n)$$

(Prove the first equality by showing that both statements $yin-x+B^complement$ and $yin(-x+B)^complement$ are equivalent with the statement $y+xnotin B$)

So that: $$B^complement=x+(-x+B^complement)inmathscr A_x$$

share|cite|improve this answer

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

add a comment | 

up vote
3
down vote

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $x+B =T_-x^-1(B)$ is Borel.

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where does it come from? Is every Borel set some countable union of open sets?
  – dEmigOd
  Jul 15 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
  – copper.hat
  Jul 15 at 18:28
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

add a comment | 

up vote
1
down vote

Let $mathscrT:=mathscrT(mathbb R^n)$ be the topology on $mathbb R^n$. It is obvious that $mathscrT$ is translation invariant, i.e. $Omegain mathscrT$ implies $Omega+ xinmathscrT$ for all $xinmathbb R^n$. Since $mathscrB:=mathscrB(mathbb R^n)$ is constructed from $mathscrT$ in a pure set theoretical way it immediately follows that $mathscrB$ is translation invariant as well. You don't have to go through the motions.

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852220%2fif-b-in-mathscrb-mathbbrn-then-x-b-in-mathscrb-mathbbrn%23new-answer', 'question_page');

);

Post as a guest

Name Email

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

If $Binmathscr A_x$ then $B=x+A$ for some $Ainmathscr B(mathbb R^n)$.

Then: $$-x+B^complement=(-x+B)^complement=A^complementinmathscr B(mathbb R^n)$$

(Prove the first equality by showing that both statements $yin-x+B^complement$ and $yin(-x+B)^complement$ are equivalent with the statement $y+xnotin B$)

So that: $$B^complement=x+(-x+B^complement)inmathscr A_x$$

share|cite|improve this answer

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

add a comment | 

up vote
3
down vote



accepted

If $Binmathscr A_x$ then $B=x+A$ for some $Ainmathscr B(mathbb R^n)$.

Then: $$-x+B^complement=(-x+B)^complement=A^complementinmathscr B(mathbb R^n)$$

(Prove the first equality by showing that both statements $yin-x+B^complement$ and $yin(-x+B)^complement$ are equivalent with the statement $y+xnotin B$)

So that: $$B^complement=x+(-x+B^complement)inmathscr A_x$$

share|cite|improve this answer

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

If $Binmathscr A_x$ then $B=x+A$ for some $Ainmathscr B(mathbb R^n)$.

Then: $$-x+B^complement=(-x+B)^complement=A^complementinmathscr B(mathbb R^n)$$

(Prove the first equality by showing that both statements $yin-x+B^complement$ and $yin(-x+B)^complement$ are equivalent with the statement $y+xnotin B$)

So that: $$B^complement=x+(-x+B^complement)inmathscr A_x$$

share|cite|improve this answer

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

If $Binmathscr A_x$ then $B=x+A$ for some $Ainmathscr B(mathbb R^n)$.

Then: $$-x+B^complement=(-x+B)^complement=A^complementinmathscr B(mathbb R^n)$$

(Prove the first equality by showing that both statements $yin-x+B^complement$ and $yin(-x+B)^complement$ are equivalent with the statement $y+xnotin B$)

So that: $$B^complement=x+(-x+B^complement)inmathscr A_x$$

share|cite|improve this answer

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 6:15

edited Jul 15 at 6:15

edited Jul 15 at 6:15

answered Jul 15 at 6:09

drhab

86.7k541118

answered Jul 15 at 6:09

drhab

86.7k541118

answered Jul 15 at 6:09

drhab

86.7k541118

86.7k541118

add a comment | 

add a comment | 

up vote
3
down vote

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $x+B =T_-x^-1(B)$ is Borel.

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where does it come from? Is every Borel set some countable union of open sets?
  – dEmigOd
  Jul 15 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
  – copper.hat
  Jul 15 at 18:28
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $x+B =T_-x^-1(B)$ is Borel.

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where does it come from? Is every Borel set some countable union of open sets?
  – dEmigOd
  Jul 15 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
  – copper.hat
  Jul 15 at 18:28
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $x+B =T_-x^-1(B)$ is Borel.

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $x+B =T_-x^-1(B)$ is Borel.

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 6:19

copper.hat

122k557156

answered Jul 15 at 6:19

copper.hat

122k557156

answered Jul 15 at 6:19

copper.hat

122k557156

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where does it come from? Is every Borel set some countable union of open sets?
  – dEmigOd
  Jul 15 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
  – copper.hat
  Jul 15 at 18:28
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where does it come from? Is every Borel set some countable union of open sets?
  – dEmigOd
  Jul 15 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
  – copper.hat
  Jul 15 at 18:28
  
  

  
  
  
  

Where does it come from? Is every Borel set some countable union of open sets?
– dEmigOd
Jul 15 at 6:32

Where does it come from? Is every Borel set some countable union of open sets?
– dEmigOd
Jul 15 at 6:32

Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
– copper.hat
Jul 15 at 18:28

Well, $ T_-x^-1(B) $ is a $sigma$-field that contains the open sets, hence the Borel sets. (And no, the set $0$ is Borel but not the union of open sets.)
– copper.hat
Jul 15 at 18:28

add a comment | 

up vote
2
down vote

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

add a comment | 

up vote
2
down vote

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

add a comment | 

up vote
2
down vote

up vote
2
down vote

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

answered Jul 15 at 6:01

Eric Wofsey

163k12189300

163k12189300

add a comment | 

add a comment | 

up vote
1
down vote

Let $mathscrT:=mathscrT(mathbb R^n)$ be the topology on $mathbb R^n$. It is obvious that $mathscrT$ is translation invariant, i.e. $Omegain mathscrT$ implies $Omega+ xinmathscrT$ for all $xinmathbb R^n$. Since $mathscrB:=mathscrB(mathbb R^n)$ is constructed from $mathscrT$ in a pure set theoretical way it immediately follows that $mathscrB$ is translation invariant as well. You don't have to go through the motions.

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

add a comment | 

up vote
1
down vote

Let $mathscrT:=mathscrT(mathbb R^n)$ be the topology on $mathbb R^n$. It is obvious that $mathscrT$ is translation invariant, i.e. $Omegain mathscrT$ implies $Omega+ xinmathscrT$ for all $xinmathbb R^n$. Since $mathscrB:=mathscrB(mathbb R^n)$ is constructed from $mathscrT$ in a pure set theoretical way it immediately follows that $mathscrB$ is translation invariant as well. You don't have to go through the motions.

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $mathscrT:=mathscrT(mathbb R^n)$ be the topology on $mathbb R^n$. It is obvious that $mathscrT$ is translation invariant, i.e. $Omegain mathscrT$ implies $Omega+ xinmathscrT$ for all $xinmathbb R^n$. Since $mathscrB:=mathscrB(mathbb R^n)$ is constructed from $mathscrT$ in a pure set theoretical way it immediately follows that $mathscrB$ is translation invariant as well. You don't have to go through the motions.

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

Let $mathscrT:=mathscrT(mathbb R^n)$ be the topology on $mathbb R^n$. It is obvious that $mathscrT$ is translation invariant, i.e. $Omegain mathscrT$ implies $Omega+ xinmathscrT$ for all $xinmathbb R^n$. Since $mathscrB:=mathscrB(mathbb R^n)$ is constructed from $mathscrT$ in a pure set theoretical way it immediately follows that $mathscrB$ is translation invariant as well. You don't have to go through the motions.

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 16:09

Christian Blatter

164k7108306

answered Jul 15 at 16:09

Christian Blatter

164k7108306

answered Jul 15 at 16:09

Christian Blatter

164k7108306

164k7108306

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852220%2fif-b-in-mathscrb-mathbbrn-then-x-b-in-mathscrb-mathbbrn%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email