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Proof that interval $]0,1[$ is an open set

Every point in the open set is a limit pointUsing the “open cover” definition of compactness to show that $[0,1]$ is compactWhat is difference between open set and open interval?Every open subset of $mathbb R$ is an $F_sigma$-set and a $G_delta$-set.Prove that interval $(0,1]$ is not compactProve: Whereas $(0,1)$ in $mathbbR$ is open or closedunion of a collection of open interval containing the rationals in $[0,1]$Every point in open set belongs to component intervalHow to prove that $L^infty([0,1]) subset L^q([0,1]) subset L^p([0,1])$?$A=Bigf in C[0,1]: f(x)neq 0,forall x in [0,1]Big$ is open in $C[0,1]$

1

$begingroup$

I have this question in my set of exercises of real analysis:

prove that the interval $]0,1[$ is an open set.

The definition says

the set $Asubset mathbbR^m$ is open if for all $a in A$, $exists delta gt 0$ such that $B[a; delta] subset A$.

Although I get the intuition of this definition somehow, I don't really know how to prove the statement mathematically. In particular, I don't know how to choose my $delta$.

Any tips are welcomed.
Thank you in advance.

real-analysis general-topology

share|cite|improve this question

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Try to prove that the complement of $]0,1[$ is closed
  $endgroup$
  – Victoria M
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The basi hint is: take a point and try to find a ball centered on it and contained in the set. I know it may sound basic, but I promise if you follow the hint you will be able to solve the problem.
  $endgroup$
  – Dog_69
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks guys for the help.
  $endgroup$
  – Mathieu Rousseau
  18 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I have this question in my set of exercises of real analysis:

prove that the interval $]0,1[$ is an open set.

The definition says

the set $Asubset mathbbR^m$ is open if for all $a in A$, $exists delta gt 0$ such that $B[a; delta] subset A$.

Although I get the intuition of this definition somehow, I don't really know how to prove the statement mathematically. In particular, I don't know how to choose my $delta$.

Any tips are welcomed.
Thank you in advance.

real-analysis general-topology

share|cite|improve this question

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Try to prove that the complement of $]0,1[$ is closed
  $endgroup$
  – Victoria M
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The basi hint is: take a point and try to find a ball centered on it and contained in the set. I know it may sound basic, but I promise if you follow the hint you will be able to solve the problem.
  $endgroup$
  – Dog_69
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks guys for the help.
  $endgroup$
  – Mathieu Rousseau
  18 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I have this question in my set of exercises of real analysis:

prove that the interval $]0,1[$ is an open set.

The definition says

the set $Asubset mathbbR^m$ is open if for all $a in A$, $exists delta gt 0$ such that $B[a; delta] subset A$.

Although I get the intuition of this definition somehow, I don't really know how to prove the statement mathematically. In particular, I don't know how to choose my $delta$.

Any tips are welcomed.
Thank you in advance.

real-analysis general-topology

share|cite|improve this question

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

J. W. Tanner

2,9541218

edited yesterday

J. W. Tanner

2,9541218

asked yesterday

Mathieu RousseauMathieu Rousseau

62

New contributor

Mathieu Rousseau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Mathieu RousseauMathieu Rousseau

62

2 Answers
2

active

oldest

votes

3

$begingroup$

If $ain A$, take $delta=mina,1-a$ (that is, $delta$ is the smallest between the distances from $a$ to $0$ and to $1$). Then $(a-delta,a+delta)subset(0,1)$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MatthieuRousseau To formally proof that last inclusion: suppose $x in (a-delta, a+delta)$. So $x > a-delta$ and as $delta le a$, we know $a-delta ge 0$ and so $x > a-delta ge 0$ hence $x >0$; also $delta le 1-a$ so $a+delta le 1$ and $x < a+delta le 1$ implies $x < 1$. Just to show we don't rely on a "picture".
  $endgroup$
  – Henno Brandsma
  21 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

For $a in (0,1)$ set $$delta = frac12(1-|2a-1|)$$

Then $B(a, delta) = (a-delta, a+delta) subseteq (0,1)$. Indeed,

$$a-delta = a-frac12(1-|2a-1|) = left(a-frac12right) + left|a-frac12right| ge 0$$

$$a+delta = a+frac12(1-|2a-1|) = left(a+frac12right) - left|a-frac12right| le left(a+frac12right) - left(a-frac12right) =1 $$

share|cite|improve this answer

answered yesterday

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

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3

$begingroup$

If $ain A$, take $delta=mina,1-a$ (that is, $delta$ is the smallest between the distances from $a$ to $0$ and to $1$). Then $(a-delta,a+delta)subset(0,1)$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MatthieuRousseau To formally proof that last inclusion: suppose $x in (a-delta, a+delta)$. So $x > a-delta$ and as $delta le a$, we know $a-delta ge 0$ and so $x > a-delta ge 0$ hence $x >0$; also $delta le 1-a$ so $a+delta le 1$ and $x < a+delta le 1$ implies $x < 1$. Just to show we don't rely on a "picture".
  $endgroup$
  – Henno Brandsma
  21 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

If $ain A$, take $delta=mina,1-a$ (that is, $delta$ is the smallest between the distances from $a$ to $0$ and to $1$). Then $(a-delta,a+delta)subset(0,1)$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MatthieuRousseau To formally proof that last inclusion: suppose $x in (a-delta, a+delta)$. So $x > a-delta$ and as $delta le a$, we know $a-delta ge 0$ and so $x > a-delta ge 0$ hence $x >0$; also $delta le 1-a$ so $a+delta le 1$ and $x < a+delta le 1$ implies $x < 1$. Just to show we don't rely on a "picture".
  $endgroup$
  – Henno Brandsma
  21 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

If $ain A$, take $delta=mina,1-a$ (that is, $delta$ is the smallest between the distances from $a$ to $0$ and to $1$). Then $(a-delta,a+delta)subset(0,1)$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

0

$begingroup$

For $a in (0,1)$ set $$delta = frac12(1-|2a-1|)$$

Then $B(a, delta) = (a-delta, a+delta) subseteq (0,1)$. Indeed,

$$a-delta = a-frac12(1-|2a-1|) = left(a-frac12right) + left|a-frac12right| ge 0$$

$$a+delta = a+frac12(1-|2a-1|) = left(a+frac12right) - left|a-frac12right| le left(a+frac12right) - left(a-frac12right) =1 $$

share|cite|improve this answer

answered yesterday

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

0

$begingroup$

For $a in (0,1)$ set $$delta = frac12(1-|2a-1|)$$

Then $B(a, delta) = (a-delta, a+delta) subseteq (0,1)$. Indeed,

$$a-delta = a-frac12(1-|2a-1|) = left(a-frac12right) + left|a-frac12right| ge 0$$

$$a+delta = a+frac12(1-|2a-1|) = left(a+frac12right) - left|a-frac12right| le left(a+frac12right) - left(a-frac12right) =1 $$

share|cite|improve this answer

answered yesterday

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

$begingroup$

For $a in (0,1)$ set $$delta = frac12(1-|2a-1|)$$

Then $B(a, delta) = (a-delta, a+delta) subseteq (0,1)$. Indeed,

$$a-delta = a-frac12(1-|2a-1|) = left(a-frac12right) + left|a-frac12right| ge 0$$

$$a+delta = a+frac12(1-|2a-1|) = left(a+frac12right) - left|a-frac12right| le left(a+frac12right) - left(a-frac12right) =1 $$

share|cite|improve this answer

answered yesterday

mechanodroidmechanodroid

28.5k62548

$endgroup$

share|cite|improve this answer

answered yesterday

mechanodroidmechanodroid

28.5k62548

answered yesterday

mechanodroidmechanodroid

28.5k62548

answered yesterday

mechanodroidmechanodroid

28.5k62548