Why is the set of all non-decreasing sequence not open in the real sequence space?

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Given the set $B=a(n)<C$ with the distance $d(a(n),b(n)):= sup|a(n)-b(n)|$.

Take the set $J:=ain B $

I know that $J$ is not open in the metric space $B$ but I want to make sure I know exactly why.

So take the null sequence $0in J$. This is an element of J as each member is zero and so the condition $a(n+1)geq a(n)$ is satisfied as zero is equal to zero.

Next consider an open ball centered at $0$, $B_r(0)$, so the taking the metric $d(0,b(n)):=sup|0-b(n)|= sup |b(n)|$ the open ball is $B_r(0)=d(0,b(n)) <r$, In other words $sup|b(n)|<r$. okay honestly after this I'm stuck. I don't know why no open ball exists in $J$ which is centered at zero . could anyone explain ?

general-topology metric-spaces

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edited Aug 2 at 14:15

feynhat

428314

asked Aug 2 at 13:52

exodius

714215

add a comment | 

up vote
2
down vote

favorite

Given the set $B=a(n)<C$ with the distance $d(a(n),b(n)):= sup|a(n)-b(n)|$.

Take the set $J:=ain B $

I know that $J$ is not open in the metric space $B$ but I want to make sure I know exactly why.

So take the null sequence $0in J$. This is an element of J as each member is zero and so the condition $a(n+1)geq a(n)$ is satisfied as zero is equal to zero.

Next consider an open ball centered at $0$, $B_r(0)$, so the taking the metric $d(0,b(n)):=sup|0-b(n)|= sup |b(n)|$ the open ball is $B_r(0)=d(0,b(n)) <r$, In other words $sup|b(n)|<r$. okay honestly after this I'm stuck. I don't know why no open ball exists in $J$ which is centered at zero . could anyone explain ?

general-topology metric-spaces

share|cite|improve this question

edited Aug 2 at 14:15

feynhat

428314

asked Aug 2 at 13:52

exodius

714215

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Given the set $B=a(n)<C$ with the distance $d(a(n),b(n)):= sup|a(n)-b(n)|$.

Take the set $J:=ain B $

I know that $J$ is not open in the metric space $B$ but I want to make sure I know exactly why.

So take the null sequence $0in J$. This is an element of J as each member is zero and so the condition $a(n+1)geq a(n)$ is satisfied as zero is equal to zero.

Next consider an open ball centered at $0$, $B_r(0)$, so the taking the metric $d(0,b(n)):=sup|0-b(n)|= sup |b(n)|$ the open ball is $B_r(0)=d(0,b(n)) <r$, In other words $sup|b(n)|<r$. okay honestly after this I'm stuck. I don't know why no open ball exists in $J$ which is centered at zero . could anyone explain ?

general-topology metric-spaces

share|cite|improve this question

edited Aug 2 at 14:15

feynhat

428314

asked Aug 2 at 13:52

exodius

714215

Given the set $B=a(n)<C$ with the distance $d(a(n),b(n)):= sup|a(n)-b(n)|$.

Take the set $J:=ain B $

I know that $J$ is not open in the metric space $B$ but I want to make sure I know exactly why.

So take the null sequence $0in J$. This is an element of J as each member is zero and so the condition $a(n+1)geq a(n)$ is satisfied as zero is equal to zero.

Next consider an open ball centered at $0$, $B_r(0)$, so the taking the metric $d(0,b(n)):=sup|0-b(n)|= sup |b(n)|$ the open ball is $B_r(0)=d(0,b(n)) <r$, In other words $sup|b(n)|<r$. okay honestly after this I'm stuck. I don't know why no open ball exists in $J$ which is centered at zero . could anyone explain ?

general-topology metric-spaces

share|cite|improve this question

edited Aug 2 at 14:15

feynhat

428314

asked Aug 2 at 13:52

exodius

714215

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 14:15

feynhat

428314

edited Aug 2 at 14:15

feynhat

428314

edited Aug 2 at 14:15

feynhat

428314

428314

asked Aug 2 at 13:52

exodius

714215

asked Aug 2 at 13:52

exodius

714215

asked Aug 2 at 13:52

exodius

714215

714215

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element
$$
c(n) := begincases
epsilon & n=1 \
0 & n ge 2.
endcases
$$
It violates the condition of $J$, but is close to zero with regard t your metric.

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
  – exodius
  Aug 2 at 14:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  could you tell me if that's correct ?
  – exodius
  Aug 2 at 14:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
  – AlgebraicsAnonymous
  Aug 2 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
  – exodius
  Aug 2 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  if not I can rewrite it again, just not sure if that's what you meant or not
  – exodius
  Aug 2 at 14:21
  

  
  
  
  

 | 
show 7 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element
$$
c(n) := begincases
epsilon & n=1 \
0 & n ge 2.
endcases
$$
It violates the condition of $J$, but is close to zero with regard t your metric.

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
  – exodius
  Aug 2 at 14:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  could you tell me if that's correct ?
  – exodius
  Aug 2 at 14:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
  – AlgebraicsAnonymous
  Aug 2 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
  – exodius
  Aug 2 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  if not I can rewrite it again, just not sure if that's what you meant or not
  – exodius
  Aug 2 at 14:21
  

  
  
  
  

 | 
show 7 more comments

up vote
2
down vote

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element
$$
c(n) := begincases
epsilon & n=1 \
0 & n ge 2.
endcases
$$
It violates the condition of $J$, but is close to zero with regard t your metric.

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
  – exodius
  Aug 2 at 14:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  could you tell me if that's correct ?
  – exodius
  Aug 2 at 14:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
  – AlgebraicsAnonymous
  Aug 2 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
  – exodius
  Aug 2 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  if not I can rewrite it again, just not sure if that's what you meant or not
  – exodius
  Aug 2 at 14:21
  

  
  
  
  

 | 
show 7 more comments

up vote
2
down vote

up vote
2
down vote

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element
$$
c(n) := begincases
epsilon & n=1 \
0 & n ge 2.
endcases
$$
It violates the condition of $J$, but is close to zero with regard t your metric.

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

The assertion is equivalent to the existence of arbitrarily small points in the complement of the set. Now consider the element
$$
c(n) := begincases
epsilon & n=1 \
0 & n ge 2.
endcases
$$
It violates the condition of $J$, but is close to zero with regard t your metric.

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

answered Aug 2 at 13:57

AlgebraicsAnonymous

66111

66111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
  – exodius
  Aug 2 at 14:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  could you tell me if that's correct ?
  – exodius
  Aug 2 at 14:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
  – AlgebraicsAnonymous
  Aug 2 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
  – exodius
  Aug 2 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  if not I can rewrite it again, just not sure if that's what you meant or not
  – exodius
  Aug 2 at 14:21
  

  
  
  
  

 | 
show 7 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
  – exodius
  Aug 2 at 14:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  could you tell me if that's correct ?
  – exodius
  Aug 2 at 14:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
  – AlgebraicsAnonymous
  Aug 2 at 14:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
  – exodius
  Aug 2 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  if not I can rewrite it again, just not sure if that's what you meant or not
  – exodius
  Aug 2 at 14:21
  

  
  
  
  

okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
– exodius
Aug 2 at 14:06

okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball $B_epsilon(0)$ but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?
– exodius
Aug 2 at 14:06

could you tell me if that's correct ?
– exodius
Aug 2 at 14:13

could you tell me if that's correct ?
– exodius
Aug 2 at 14:13

Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
– AlgebraicsAnonymous
Aug 2 at 14:15

Unfortunately my browser does not render comments with math-dollars. Could you rephrase accordingly?
– AlgebraicsAnonymous
Aug 2 at 14:15

okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
– exodius
Aug 2 at 14:16

okay I think I understand...c(n) is arbitrarily close to my null sequence 0 so it exists in an open ball Bϵ(0) but it does not exist in Jso then this open ball does not exist soley in J and so J cannot be open ?I took out the math dollars around the open ball... i thjink this is what you meant
– exodius
Aug 2 at 14:16

if not I can rewrite it again, just not sure if that's what you meant or not
– exodius
Aug 2 at 14:21

if not I can rewrite it again, just not sure if that's what you meant or not
– exodius
Aug 2 at 14:21

 | 
show 7 more comments

 

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