Does every point in a metric space have another point a rational distance away?

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Is the following statement true or false?

Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.

I'm not able to find a counter example.

general-topology metric-spaces

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edited 52 mins ago

Nate Eldredge

61.1k579165

asked 1 hour ago

jasmine

1,258314

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What have you tried for finding counterexamples?
  – Dark Malthorp
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Think about the discrete metric, but choose an irrational number instead of $1$.
  – Richard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Richard good logics
  – jasmine
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NateEldredge okkss
  – jasmine
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Is the following statement true or false?

Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.

I'm not able to find a counter example.

general-topology metric-spaces

share|cite|improve this question

edited 52 mins ago

Nate Eldredge

61.1k579165

asked 1 hour ago

jasmine

1,258314

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What have you tried for finding counterexamples?
  – Dark Malthorp
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Think about the discrete metric, but choose an irrational number instead of $1$.
  – Richard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Richard good logics
  – jasmine
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NateEldredge okkss
  – jasmine
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Is the following statement true or false?

Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.

I'm not able to find a counter example.

general-topology metric-spaces

share|cite|improve this question

edited 52 mins ago

Nate Eldredge

61.1k579165

asked 1 hour ago

jasmine

1,258314

Is the following statement true or false?

Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.

I'm not able to find a counter example.

general-topology metric-spaces

general-topology metric-spaces

share|cite|improve this question

edited 52 mins ago

Nate Eldredge

61.1k579165

asked 1 hour ago

jasmine

1,258314

share|cite|improve this question

edited 52 mins ago

Nate Eldredge

61.1k579165

asked 1 hour ago

jasmine

1,258314

share|cite|improve this question

share|cite|improve this question

edited 52 mins ago

Nate Eldredge

61.1k579165

edited 52 mins ago

Nate Eldredge

61.1k579165

edited 52 mins ago

Nate Eldredge

61.1k579165

61.1k579165

asked 1 hour ago

jasmine

1,258314

asked 1 hour ago

jasmine

1,258314

asked 1 hour ago

jasmine

1,258314

1,258314

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What have you tried for finding counterexamples?
  – Dark Malthorp
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Think about the discrete metric, but choose an irrational number instead of $1$.
  – Richard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Richard good logics
  – jasmine
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NateEldredge okkss
  – jasmine
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What have you tried for finding counterexamples?
  – Dark Malthorp
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Think about the discrete metric, but choose an irrational number instead of $1$.
  – Richard
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Richard good logics
  – jasmine
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NateEldredge okkss
  – jasmine
  1 hour ago
  

  
  
  
  

2

2

What have you tried for finding counterexamples?
– Dark Malthorp
1 hour ago

What have you tried for finding counterexamples?
– Dark Malthorp
1 hour ago

4

4

Think about the discrete metric, but choose an irrational number instead of $1$.
– Richard
1 hour ago

Think about the discrete metric, but choose an irrational number instead of $1$.
– Richard
1 hour ago

@Richard good logics
– jasmine
1 hour ago

@Richard good logics
– jasmine
1 hour ago

@NateEldredge okkss
– jasmine
1 hour ago

@NateEldredge okkss
– jasmine
1 hour ago

add a comment | 

3 Answers
3

active

oldest

votes

up vote
4
down vote



accepted

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Alternatively, take $X = 0$ with trivial metric.
  – ConMan
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

add a comment | 

up vote
2
down vote

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

share|cite|improve this answer

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very clever example.
  – Matt Samuel
  36 mins ago
  

  
  
  
  

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



accepted

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Alternatively, take $X = 0$ with trivial metric.
  – ConMan
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Alternatively, take $X = 0$ with trivial metric.
  – ConMan
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Noah Riggenbach

63716

answered 1 hour ago

Noah Riggenbach

63716

answered 1 hour ago

Noah Riggenbach

63716

63716

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Alternatively, take $X = 0$ with trivial metric.
  – ConMan
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Alternatively, take $X = 0$ with trivial metric.
  – ConMan
  1 hour ago
  

  
  
  
  

2

2

Alternatively, take $X = 0$ with trivial metric.
– ConMan
1 hour ago

Alternatively, take $X = 0$ with trivial metric.
– ConMan
1 hour ago

add a comment | 

up vote
2
down vote

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

add a comment | 

up vote
2
down vote

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

add a comment | 

up vote
2
down vote

up vote
2
down vote

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Mjiig

34727

answered 1 hour ago

Mjiig

34727

answered 1 hour ago

Mjiig

34727

34727

add a comment | 

add a comment | 

up vote
2
down vote

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

share|cite|improve this answer

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very clever example.
  – Matt Samuel
  36 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

share|cite|improve this answer

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very clever example.
  – Matt Samuel
  36 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

share|cite|improve this answer

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

share|cite|improve this answer

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

share|cite|improve this answer

share|cite|improve this answer

edited 24 mins ago

edited 24 mins ago

edited 24 mins ago

answered 1 hour ago

Nate Eldredge

61.1k579165

answered 1 hour ago

Nate Eldredge

61.1k579165

answered 1 hour ago

Nate Eldredge

61.1k579165

61.1k579165

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very clever example.
  – Matt Samuel
  36 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very clever example.
  – Matt Samuel
  36 mins ago
  

  
  
  
  

Very clever example.
– Matt Samuel
36 mins ago

Very clever example.
– Matt Samuel
36 mins ago

add a comment | 

 

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