about $C([0,1])$ with sup metric

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Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] mapsto Bbb R$ with $C([0,1])$ $$ d(f,g) = sup _x in [0,1]f(x)-g(x) , $$
so let $$A= leftf in C([0,1]) middle$$ now is $A$ open , close , bounded , connect or compact ?

I think $A$ is open because for every $f in A $ we have $B_t (f) subseteq A $ such that $t:= 1- int_0^1 f(x) mathrmdx $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= frac1n$ then $ 0< int_0^1 f_n(x) mathrmdx=frac1n <1 $ and for every $1< n in mathbbN$ ,$ f_n(x) in A$ and $lim_n to infty f_n(x)=0$ then $ int_0^1 lim_n to infty f_n(x) mathrmdx=0 $ then
$ lim_n to infty f_n(x) notin A$ hence $A$ is not close and yet $A$ is not compact .

general-topology metric-spaces compactness normed-spaces connectedness

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edited Jul 23 at 15:13

José Carlos Santos

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asked Jul 23 at 15:02

amir bahadory

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up vote
4
down vote

favorite

1

Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] mapsto Bbb R$ with $C([0,1])$ $$ d(f,g) = sup _x in [0,1]f(x)-g(x) , $$
so let $$A= leftf in C([0,1]) middle$$ now is $A$ open , close , bounded , connect or compact ?

I think $A$ is open because for every $f in A $ we have $B_t (f) subseteq A $ such that $t:= 1- int_0^1 f(x) mathrmdx $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= frac1n$ then $ 0< int_0^1 f_n(x) mathrmdx=frac1n <1 $ and for every $1< n in mathbbN$ ,$ f_n(x) in A$ and $lim_n to infty f_n(x)=0$ then $ int_0^1 lim_n to infty f_n(x) mathrmdx=0 $ then
$ lim_n to infty f_n(x) notin A$ hence $A$ is not close and yet $A$ is not compact .

general-topology metric-spaces compactness normed-spaces connectedness

share|cite|improve this question

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

asked Jul 23 at 15:02

amir bahadory

1,238317

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] mapsto Bbb R$ with $C([0,1])$ $$ d(f,g) = sup _x in [0,1]f(x)-g(x) , $$
so let $$A= leftf in C([0,1]) middle$$ now is $A$ open , close , bounded , connect or compact ?

I think $A$ is open because for every $f in A $ we have $B_t (f) subseteq A $ such that $t:= 1- int_0^1 f(x) mathrmdx $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= frac1n$ then $ 0< int_0^1 f_n(x) mathrmdx=frac1n <1 $ and for every $1< n in mathbbN$ ,$ f_n(x) in A$ and $lim_n to infty f_n(x)=0$ then $ int_0^1 lim_n to infty f_n(x) mathrmdx=0 $ then
$ lim_n to infty f_n(x) notin A$ hence $A$ is not close and yet $A$ is not compact .

general-topology metric-spaces compactness normed-spaces connectedness

share|cite|improve this question

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

asked Jul 23 at 15:02

amir bahadory

1,238317

Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] mapsto Bbb R$ with $C([0,1])$ $$ d(f,g) = sup _x in [0,1]f(x)-g(x) , $$
so let $$A= leftf in C([0,1]) middle$$ now is $A$ open , close , bounded , connect or compact ?

I think $A$ is open because for every $f in A $ we have $B_t (f) subseteq A $ such that $t:= 1- int_0^1 f(x) mathrmdx $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= frac1n$ then $ 0< int_0^1 f_n(x) mathrmdx=frac1n <1 $ and for every $1< n in mathbbN$ ,$ f_n(x) in A$ and $lim_n to infty f_n(x)=0$ then $ int_0^1 lim_n to infty f_n(x) mathrmdx=0 $ then
$ lim_n to infty f_n(x) notin A$ hence $A$ is not close and yet $A$ is not compact .

general-topology metric-spaces compactness normed-spaces connectedness

share|cite|improve this question

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

asked Jul 23 at 15:02

amir bahadory

1,238317

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

edited Jul 23 at 15:13

José Carlos Santos

113k1698176

113k1698176

asked Jul 23 at 15:02

amir bahadory

1,238317

asked Jul 23 at 15:02

amir bahadory

1,238317

asked Jul 23 at 15:02

amir bahadory

1,238317

1,238317

add a comment | 

add a comment | 

3 Answers
3

active

oldest

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up vote
4
down vote

The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that

$A=(int_0^1dx)^-1((0,1))$

that it is open because it is the inverse image of an open set with respect to a continuos function.

So $A$ it is not closed because $Aneq C([0,1])$, $Aneq emptyset$ and $C([0,1])$ is a topological vector space and so it is connected.

$A$ it is not compact because it is not closed in $C([0,1])$

It is connected because it is a convex subset.

It is not bounded because you can choose for every $epsilon>0$ a continuos function $f_epsilon$ such that there exists $xin [0,1]$ for which $f_epsilon(x)>epsilon$ but $0<int_0^1f_epsilon(x)dx<1$

share|cite|improve this answer

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
  – DanielWainfleet
  Jul 27 at 9:20
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

$A$ is not bounded

The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.

$A$ is connected

$A$ is connected because it is convex.

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edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

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The map$$beginarrayrcccIcolon&Cbigl([0,1]bigr)&longrightarrow&mathbbR\&f&mapsto&displaystyleint_0^1f(x),mathrm dxendarray$$is continuous. Since $A=I^-1bigl((0,1)bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,gin A$, just consider$$beginarrayrcccgammacolon&[0,1]&longrightarrow&A\&t&mapsto&tg+(1-t)f.endarray$$Then $gamma$ is continuous, $gamma(0)=f$, and $gamma(1)=g$. Therefore $A$ is path-connected.

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $A$ bounded ?
  – amir bahadory
  Jul 23 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, since $Asubset D_1(0)$.
  – José Carlos Santos
  Jul 23 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why $A subset D_1(0) $ ?
  – amir bahadory
  Jul 23 at 15:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JoséCarlosSantos José, $A$ is not bounded. See my answer.
  – mathcounterexamples.net
  Jul 23 at 15:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net You're right. My mistake.
  – José Carlos Santos
  Jul 23 at 15:24
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that

$A=(int_0^1dx)^-1((0,1))$

that it is open because it is the inverse image of an open set with respect to a continuos function.

So $A$ it is not closed because $Aneq C([0,1])$, $Aneq emptyset$ and $C([0,1])$ is a topological vector space and so it is connected.

$A$ it is not compact because it is not closed in $C([0,1])$

It is connected because it is a convex subset.

It is not bounded because you can choose for every $epsilon>0$ a continuos function $f_epsilon$ such that there exists $xin [0,1]$ for which $f_epsilon(x)>epsilon$ but $0<int_0^1f_epsilon(x)dx<1$

share|cite|improve this answer

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
  – DanielWainfleet
  Jul 27 at 9:20
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that

$A=(int_0^1dx)^-1((0,1))$

that it is open because it is the inverse image of an open set with respect to a continuos function.

So $A$ it is not closed because $Aneq C([0,1])$, $Aneq emptyset$ and $C([0,1])$ is a topological vector space and so it is connected.

$A$ it is not compact because it is not closed in $C([0,1])$

It is connected because it is a convex subset.

It is not bounded because you can choose for every $epsilon>0$ a continuos function $f_epsilon$ such that there exists $xin [0,1]$ for which $f_epsilon(x)>epsilon$ but $0<int_0^1f_epsilon(x)dx<1$

share|cite|improve this answer

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
  – DanielWainfleet
  Jul 27 at 9:20
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that

$A=(int_0^1dx)^-1((0,1))$

that it is open because it is the inverse image of an open set with respect to a continuos function.

So $A$ it is not closed because $Aneq C([0,1])$, $Aneq emptyset$ and $C([0,1])$ is a topological vector space and so it is connected.

$A$ it is not compact because it is not closed in $C([0,1])$

It is connected because it is a convex subset.

It is not bounded because you can choose for every $epsilon>0$ a continuos function $f_epsilon$ such that there exists $xin [0,1]$ for which $f_epsilon(x)>epsilon$ but $0<int_0^1f_epsilon(x)dx<1$

share|cite|improve this answer

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that

$A=(int_0^1dx)^-1((0,1))$

that it is open because it is the inverse image of an open set with respect to a continuos function.

So $A$ it is not closed because $Aneq C([0,1])$, $Aneq emptyset$ and $C([0,1])$ is a topological vector space and so it is connected.

$A$ it is not compact because it is not closed in $C([0,1])$

It is connected because it is a convex subset.

It is not bounded because you can choose for every $epsilon>0$ a continuos function $f_epsilon$ such that there exists $xin [0,1]$ for which $f_epsilon(x)>epsilon$ but $0<int_0^1f_epsilon(x)dx<1$

share|cite|improve this answer

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 15:28

edited Jul 23 at 15:28

edited Jul 23 at 15:28

answered Jul 23 at 15:05

Federico Fallucca

1,42118

answered Jul 23 at 15:05

Federico Fallucca

1,42118

answered Jul 23 at 15:05

Federico Fallucca

1,42118

1,42118

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
  – DanielWainfleet
  Jul 27 at 9:20
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
  – DanielWainfleet
  Jul 27 at 9:20
  
  

  
  
  
  

1

1

We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
– DanielWainfleet
Jul 27 at 9:20

We can also say that $A$ is not compact because it is unbounded. Also that if $f_n(x)=1/n$ for all $xin [0,1]$ then the set $B=f_n:nin Bbb N subset A,$ and the constant function $0$ is in $overline B$ so $A$ is not closed...........+1
– DanielWainfleet
Jul 27 at 9:20

add a comment | 

up vote
3
down vote

$A$ is not bounded

The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.

$A$ is connected

$A$ is connected because it is convex.

share|cite|improve this answer

edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

add a comment | 

up vote
3
down vote

$A$ is not bounded

The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.

$A$ is connected

$A$ is connected because it is convex.

share|cite|improve this answer

edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

add a comment | 

up vote
3
down vote

up vote
3
down vote

$A$ is not bounded

The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.

$A$ is connected

$A$ is connected because it is convex.

share|cite|improve this answer

edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

$A$ is not bounded

The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.

$A$ is connected

$A$ is connected because it is convex.

share|cite|improve this answer

edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 15:30

edited Jul 23 at 15:30

edited Jul 23 at 15:30

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

answered Jul 23 at 15:22

mathcounterexamples.net

23.9k21653

23.9k21653

add a comment | 

add a comment | 

up vote
0
down vote

The map$$beginarrayrcccIcolon&Cbigl([0,1]bigr)&longrightarrow&mathbbR\&f&mapsto&displaystyleint_0^1f(x),mathrm dxendarray$$is continuous. Since $A=I^-1bigl((0,1)bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,gin A$, just consider$$beginarrayrcccgammacolon&[0,1]&longrightarrow&A\&t&mapsto&tg+(1-t)f.endarray$$Then $gamma$ is continuous, $gamma(0)=f$, and $gamma(1)=g$. Therefore $A$ is path-connected.

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $A$ bounded ?
  – amir bahadory
  Jul 23 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, since $Asubset D_1(0)$.
  – José Carlos Santos
  Jul 23 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why $A subset D_1(0) $ ?
  – amir bahadory
  Jul 23 at 15:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JoséCarlosSantos José, $A$ is not bounded. See my answer.
  – mathcounterexamples.net
  Jul 23 at 15:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net You're right. My mistake.
  – José Carlos Santos
  Jul 23 at 15:24
  

  
  
  
  

add a comment | 

up vote
0
down vote

The map$$beginarrayrcccIcolon&Cbigl([0,1]bigr)&longrightarrow&mathbbR\&f&mapsto&displaystyleint_0^1f(x),mathrm dxendarray$$is continuous. Since $A=I^-1bigl((0,1)bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,gin A$, just consider$$beginarrayrcccgammacolon&[0,1]&longrightarrow&A\&t&mapsto&tg+(1-t)f.endarray$$Then $gamma$ is continuous, $gamma(0)=f$, and $gamma(1)=g$. Therefore $A$ is path-connected.

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $A$ bounded ?
  – amir bahadory
  Jul 23 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, since $Asubset D_1(0)$.
  – José Carlos Santos
  Jul 23 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why $A subset D_1(0) $ ?
  – amir bahadory
  Jul 23 at 15:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JoséCarlosSantos José, $A$ is not bounded. See my answer.
  – mathcounterexamples.net
  Jul 23 at 15:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net You're right. My mistake.
  – José Carlos Santos
  Jul 23 at 15:24
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The map$$beginarrayrcccIcolon&Cbigl([0,1]bigr)&longrightarrow&mathbbR\&f&mapsto&displaystyleint_0^1f(x),mathrm dxendarray$$is continuous. Since $A=I^-1bigl((0,1)bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,gin A$, just consider$$beginarrayrcccgammacolon&[0,1]&longrightarrow&A\&t&mapsto&tg+(1-t)f.endarray$$Then $gamma$ is continuous, $gamma(0)=f$, and $gamma(1)=g$. Therefore $A$ is path-connected.

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

The map$$beginarrayrcccIcolon&Cbigl([0,1]bigr)&longrightarrow&mathbbR\&f&mapsto&displaystyleint_0^1f(x),mathrm dxendarray$$is continuous. Since $A=I^-1bigl((0,1)bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,gin A$, just consider$$beginarrayrcccgammacolon&[0,1]&longrightarrow&A\&t&mapsto&tg+(1-t)f.endarray$$Then $gamma$ is continuous, $gamma(0)=f$, and $gamma(1)=g$. Therefore $A$ is path-connected.

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

answered Jul 23 at 15:12

José Carlos Santos

113k1698176

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $A$ bounded ?
  – amir bahadory
  Jul 23 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, since $Asubset D_1(0)$.
  – José Carlos Santos
  Jul 23 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why $A subset D_1(0) $ ?
  – amir bahadory
  Jul 23 at 15:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JoséCarlosSantos José, $A$ is not bounded. See my answer.
  – mathcounterexamples.net
  Jul 23 at 15:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net You're right. My mistake.
  – José Carlos Santos
  Jul 23 at 15:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is $A$ bounded ?
  – amir bahadory
  Jul 23 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, since $Asubset D_1(0)$.
  – José Carlos Santos
  Jul 23 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why $A subset D_1(0) $ ?
  – amir bahadory
  Jul 23 at 15:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JoséCarlosSantos José, $A$ is not bounded. See my answer.
  – mathcounterexamples.net
  Jul 23 at 15:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net You're right. My mistake.
  – José Carlos Santos
  Jul 23 at 15:24
  

  
  
  
  

Is $A$ bounded ?
– amir bahadory
Jul 23 at 15:14

Is $A$ bounded ?
– amir bahadory
Jul 23 at 15:14

Yes, since $Asubset D_1(0)$.
– José Carlos Santos
Jul 23 at 15:15

Yes, since $Asubset D_1(0)$.
– José Carlos Santos
Jul 23 at 15:15

Why $A subset D_1(0) $ ?
– amir bahadory
Jul 23 at 15:20

Why $A subset D_1(0) $ ?
– amir bahadory
Jul 23 at 15:20

1

1

@JoséCarlosSantos José, $A$ is not bounded. See my answer.
– mathcounterexamples.net
Jul 23 at 15:23

@JoséCarlosSantos José, $A$ is not bounded. See my answer.
– mathcounterexamples.net
Jul 23 at 15:23

@mathcounterexamples.net You're right. My mistake.
– José Carlos Santos
Jul 23 at 15:24

@mathcounterexamples.net You're right. My mistake.
– José Carlos Santos
Jul 23 at 15:24

add a comment | 

 

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