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Uncountable collection of open sets.

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Can there be an uncountable collection of open sets in $Bbb R^n$?

My idea: Since every open set contains at least one rational number, I can match each open sets to rational numbers and rational numbers are countable. Thus collection of open sets must be countable.

My professor said "the union of an arbitrary collection of open subsets of $M$ is open, and arbitrary mean either finite, countably infinite or uncountably infinite".
Here $M$ is a metric space.

real-analysis metric-spaces

share|cite|improve this question

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

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  The issue with your explanation is that there is no guarantee there is a unique rational for each open set.
  $endgroup$
  – Don Thousand
  Mar 12 at 4:22
  

  
  
  
  


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  @DonThousand Do you mean each open sets should have one rational numbers or there must be no two open sets that have same rational numbers? Sorry im not good at english
  $endgroup$
  – un Han
  Mar 12 at 4:33
  

  
  
  
  


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  If you want just an uncountable collection of open sets in $mathbb{R}^n$, consider this set $mathscr{F} = leftlbrace B_1 left( x right) | x in mathbb{R}^n rightrbrace$, where $B_1 left( x right) = leftlbrace y in mathbb{R}^n | d left( x, y right) < 1 rightrbrace$ is the open ball of radius $1$ centered at $x$ in $mathbb{R}^n$. This is an uncountable collection of open sets.
  $endgroup$
  – Aniruddha Deshmukh
  Mar 12 at 4:41
  
  
  

  
  
  
  


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  You have proved indeed that there is no uncountable collection of pairwise disjoint open sets in $mathbb{R}^n$.
  $endgroup$
  – user539887
  Mar 12 at 10:01
  

  
  
  
  

add a comment | 

4

1

$begingroup$

Can there be an uncountable collection of open sets in $Bbb R^n$?

My idea: Since every open set contains at least one rational number, I can match each open sets to rational numbers and rational numbers are countable. Thus collection of open sets must be countable.

My professor said "the union of an arbitrary collection of open subsets of $M$ is open, and arbitrary mean either finite, countably infinite or uncountably infinite".
Here $M$ is a metric space.

real-analysis metric-spaces

share|cite|improve this question

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The issue with your explanation is that there is no guarantee there is a unique rational for each open set.
  $endgroup$
  – Don Thousand
  Mar 12 at 4:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand Do you mean each open sets should have one rational numbers or there must be no two open sets that have same rational numbers? Sorry im not good at english
  $endgroup$
  – un Han
  Mar 12 at 4:33
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If you want just an uncountable collection of open sets in $mathbb{R}^n$, consider this set $mathscr{F} = leftlbrace B_1 left( x right) | x in mathbb{R}^n rightrbrace$, where $B_1 left( x right) = leftlbrace y in mathbb{R}^n | d left( x, y right) < 1 rightrbrace$ is the open ball of radius $1$ centered at $x$ in $mathbb{R}^n$. This is an uncountable collection of open sets.
  $endgroup$
  – Aniruddha Deshmukh
  Mar 12 at 4:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have proved indeed that there is no uncountable collection of pairwise disjoint open sets in $mathbb{R}^n$.
  $endgroup$
  – user539887
  Mar 12 at 10:01
  

  
  
  
  

add a comment | 

$begingroup$

Can there be an uncountable collection of open sets in $Bbb R^n$?

My idea: Since every open set contains at least one rational number, I can match each open sets to rational numbers and rational numbers are countable. Thus collection of open sets must be countable.

My professor said "the union of an arbitrary collection of open subsets of $M$ is open, and arbitrary mean either finite, countably infinite or uncountably infinite".
Here $M$ is a metric space.

real-analysis metric-spaces

share|cite|improve this question

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

$endgroup$

share|cite|improve this question

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

share|cite|improve this question

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

edited Mar 12 at 4:34

Thomas Shelby

4,1842726

asked Mar 12 at 4:19

un Hanun Han

241

asked Mar 12 at 4:19

un Hanun Han

241