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Non-deterministic Finite Automata | Sipser Example 1.16

Show that a compact metric space $(X,d)$ has countably many clopen sets.

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1

$begingroup$

Let $(X, d)$ be a compact, metric space. Then in $X$ there is at most countable amount of clopen sets. Hint: Compact metric spaces are separable.

That was an optional, harder problem to solve on my introductory topology course exam, which I had no idea how to prove.

general-topology metric-spaces

share|cite|improve this question

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are non-separable compact spaces.
  $endgroup$
  – José Carlos Santos
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $(X, d)$ be a compact, metric space. Then in $X$ there is at most countable amount of clopen sets. Hint: Compact metric spaces are separable.

That was an optional, harder problem to solve on my introductory topology course exam, which I had no idea how to prove.

general-topology metric-spaces

share|cite|improve this question

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are non-separable compact spaces.
  $endgroup$
  – José Carlos Santos
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $(X, d)$ be a compact, metric space. Then in $X$ there is at most countable amount of clopen sets. Hint: Compact metric spaces are separable.

That was an optional, harder problem to solve on my introductory topology course exam, which I had no idea how to prove.

general-topology metric-spaces

share|cite|improve this question

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

$endgroup$

share|cite|improve this question

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

share|cite|improve this question

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

edited 3 mins ago

user21820

40.7k545165

edited 3 mins ago

user21820

40.7k545165

asked 4 hours ago

math_beginnermath_beginner

1828

asked 4 hours ago

math_beginnermath_beginner

1828

1 Answer
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4

$begingroup$

This is false in general (it holds in metric spaces though).

General counterexample: ${0,1}^I$ in the product topology has $|I|$ many clopen sets (all sets of the form $pi_i^{-1}[{j}]$ for $i in I, j=0,1$ for starters).

In metric spaces, let $mathcal{B}$ be a countable base for $X$ (any compact metric space is separable and hence has a countable base). If $C$ is clopen, for each $x in C$ we can pick $B_x$ in the base such that $x in B_x subseteq C$ by openness of $C$. $C$ is also closed and hence compact so finitely many basic element also cover $C$.

So every clopen $C$ we can write as a finite union of basic sets from a countable base. There are only countably many such finite unions.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My fault - X was indeed a metric space. I have edited my question.
  $endgroup$
  – math_beginner
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math_beginner I expanded my answer.
  $endgroup$
  – Henno Brandsma
  3 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

This is false in general (it holds in metric spaces though).

General counterexample: ${0,1}^I$ in the product topology has $|I|$ many clopen sets (all sets of the form $pi_i^{-1}[{j}]$ for $i in I, j=0,1$ for starters).

In metric spaces, let $mathcal{B}$ be a countable base for $X$ (any compact metric space is separable and hence has a countable base). If $C$ is clopen, for each $x in C$ we can pick $B_x$ in the base such that $x in B_x subseteq C$ by openness of $C$. $C$ is also closed and hence compact so finitely many basic element also cover $C$.

So every clopen $C$ we can write as a finite union of basic sets from a countable base. There are only countably many such finite unions.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My fault - X was indeed a metric space. I have edited my question.
  $endgroup$
  – math_beginner
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math_beginner I expanded my answer.
  $endgroup$
  – Henno Brandsma
  3 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

This is false in general (it holds in metric spaces though).

General counterexample: ${0,1}^I$ in the product topology has $|I|$ many clopen sets (all sets of the form $pi_i^{-1}[{j}]$ for $i in I, j=0,1$ for starters).

In metric spaces, let $mathcal{B}$ be a countable base for $X$ (any compact metric space is separable and hence has a countable base). If $C$ is clopen, for each $x in C$ we can pick $B_x$ in the base such that $x in B_x subseteq C$ by openness of $C$. $C$ is also closed and hence compact so finitely many basic element also cover $C$.

So every clopen $C$ we can write as a finite union of basic sets from a countable base. There are only countably many such finite unions.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My fault - X was indeed a metric space. I have edited my question.
  $endgroup$
  – math_beginner
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math_beginner I expanded my answer.
  $endgroup$
  – Henno Brandsma
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is false in general (it holds in metric spaces though).

General counterexample: ${0,1}^I$ in the product topology has $|I|$ many clopen sets (all sets of the form $pi_i^{-1}[{j}]$ for $i in I, j=0,1$ for starters).

In metric spaces, let $mathcal{B}$ be a countable base for $X$ (any compact metric space is separable and hence has a countable base). If $C$ is clopen, for each $x in C$ we can pick $B_x$ in the base such that $x in B_x subseteq C$ by openness of $C$. $C$ is also closed and hence compact so finitely many basic element also cover $C$.

So every clopen $C$ we can write as a finite union of basic sets from a countable base. There are only countably many such finite unions.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130

answered 4 hours ago

Henno BrandsmaHenno Brandsma

119k351130