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Does every point in a metric space have another point a rational distance away?
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1
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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
edited 52 mins ago
Nate Eldredge
61.1k579165
asked 1 hour ago
jasmine
1,258314
add a comment |Â
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
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 |Â
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
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
edited 52 mins ago
Nate Eldredge
61.1k579165
asked 1 hour ago
jasmine
1,258314
edited 52 mins ago
Nate Eldredge
61.1k579165
asked 1 hour ago
jasmine
1,258314
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 |Â
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 |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.
answered 1 hour ago
Noah Riggenbach
63716
2
Alternatively, take $X = 0$ with trivial metric.
– ConMan
1 hour ago
add a comment |Â
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.
answered 1 hour ago
Mjiig
34727
add a comment |Â
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.)
answered 1 hour ago
Nate Eldredge
61.1k579165
Very clever example.
– Matt Samuel
36 mins ago
add a comment |Â
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 $xneq y$.
answered 1 hour ago
Noah Riggenbach
63716
2
Alternatively, take $X = 0$ with trivial metric.
– ConMan
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.
answered 1 hour ago
Noah Riggenbach
63716
2
Alternatively, take $X = 0$ with trivial metric.
– ConMan
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.
answered 1 hour ago
Noah Riggenbach
63716
False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.
answered 1 hour ago
Noah Riggenbach
63716
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 |Â
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 |Â
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.
answered 1 hour ago
Mjiig
34727
add a comment |Â
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.
answered 1 hour ago
Mjiig
34727
add a comment |Â
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.
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.
answered 1 hour ago
Mjiig
34727
answered 1 hour ago
Mjiig
34727
answered 1 hour ago
Mjiig
34727
answered 1 hour ago
Mjiig
34727
34727
add a comment |Â
add a comment |Â
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.)
answered 1 hour ago
Nate Eldredge
61.1k579165
Very clever example.
– Matt Samuel
36 mins ago
add a comment |Â
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.)
answered 1 hour ago
Nate Eldredge
61.1k579165
Very clever example.
– Matt Samuel
36 mins ago
add a comment |Â
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.)
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.)
answered 1 hour ago
Nate Eldredge
61.1k579165
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 |Â
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 |Â
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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