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Uncountable set with a non-discrete metric

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Uncountable set with a non-discrete metric

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2

$begingroup$

Let $X$ be an uncountable set. Can we always define a metric on this set such that $X$ has a metric topology which is not discrete?

general-topology metric-spaces

share|cite|improve this question

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

$endgroup$

add a comment | 

2

$begingroup$

Let $X$ be an uncountable set. Can we always define a metric on this set such that $X$ has a metric topology which is not discrete?

general-topology metric-spaces

share|cite|improve this question

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

$endgroup$

add a comment | 

$begingroup$

Let $X$ be an uncountable set. Can we always define a metric on this set such that $X$ has a metric topology which is not discrete?

general-topology metric-spaces

share|cite|improve this question

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

$endgroup$

share|cite|improve this question

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

share|cite|improve this question

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

asked 35 mins ago

Ajay Kumar NairAjay Kumar Nair

407

3 Answers
3

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oldest

votes

4

$begingroup$

Yes. Let $X^star={x_0,x_1,x_2,ldots}$ be a countable subset of $X$ and consider the distance $d$ in $X$ such that the distance between any two distinct points is $1$ when at least one of them is outside $X^star$ and such that, if $m,ninmathbb{Z}^+$ and $mneq n$, then$$d(x_m,x_n)=begin{cases}leftlvertfrac1m-frac1nrightrvert&text{ if }m,nneq0\frac1m&text{ if }n=0\frac1n&text{ if }m=0.end{cases}$$Then $x_0=lim_{ntoinfty}x_n$ and therefore this metric doesn't induce the discrete topology.

share|cite|improve this answer

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

$endgroup$

add a comment | 

4

$begingroup$

Of course, take any nondiscrete metric space, say $Y$ and embed it into $X$ (this assumes $Y$ has cardinality lesser then $X$). This can be $mathbb{Q}$ or $mathbb{R}$ for example. You modify the metric a bit by using $d(x,y)=min(d_Y(x,y),1)$. So you now have nondiscrete metric for elements in $Ysubseteq X$. For any other pair define the distance as $1$ unless they are equal (which obviously has to be $0$).

share|cite|improve this answer

edited 19 mins ago

answered 24 mins ago

freakishfreakish

12.5k1630

$endgroup$

add a comment | 

1

$begingroup$

If $X$ has cardinality $mathfrak{c}$, then the answer is yes: suppose that $phi : X to (0,1)$ is a bijection. Then define $d(x,y) = |phi(x) - phi(y)|$ for all $x,y in X$. This puts a topology on $X$ that is not discrete, since the usual topology on $Bbb{R}$ is not discrete.

If $X$ has cardinality greater than $mathfrak{c}$, then the answer is again yes because you can "cheat" by doing the following: choose a subset $U$ of $X$ of cardinality $mathfrak{c}$, and for all $x,y in U$ define (as before) $d(x,y) = |phi(x) - phi(y)|$. Put the discrete metric on $V = X setminus U$, that is, for all $x,y in X setminus U$, define $d(x,y) = 1$ if $xneq y$ and $d(x,y) =0$ if $x=y$. And finally, define $d(x,y) = 1$ if $x in U$, and $y in V$ (or $x in V$ and $y in U$).

Then, one can check that the triangle inequality still holds. The topology this metric induces is not discrete because of the way we have defined the metric on $U subset X$.

share|cite|improve this answer

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

$endgroup$

add a comment | 

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4

$begingroup$

Yes. Let $X^star={x_0,x_1,x_2,ldots}$ be a countable subset of $X$ and consider the distance $d$ in $X$ such that the distance between any two distinct points is $1$ when at least one of them is outside $X^star$ and such that, if $m,ninmathbb{Z}^+$ and $mneq n$, then$$d(x_m,x_n)=begin{cases}leftlvertfrac1m-frac1nrightrvert&text{ if }m,nneq0\frac1m&text{ if }n=0\frac1n&text{ if }m=0.end{cases}$$Then $x_0=lim_{ntoinfty}x_n$ and therefore this metric doesn't induce the discrete topology.

share|cite|improve this answer

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

$endgroup$

add a comment | 

4

$begingroup$

Yes. Let $X^star={x_0,x_1,x_2,ldots}$ be a countable subset of $X$ and consider the distance $d$ in $X$ such that the distance between any two distinct points is $1$ when at least one of them is outside $X^star$ and such that, if $m,ninmathbb{Z}^+$ and $mneq n$, then$$d(x_m,x_n)=begin{cases}leftlvertfrac1m-frac1nrightrvert&text{ if }m,nneq0\frac1m&text{ if }n=0\frac1n&text{ if }m=0.end{cases}$$Then $x_0=lim_{ntoinfty}x_n$ and therefore this metric doesn't induce the discrete topology.

share|cite|improve this answer

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

$endgroup$

add a comment | 

$begingroup$

Yes. Let $X^star={x_0,x_1,x_2,ldots}$ be a countable subset of $X$ and consider the distance $d$ in $X$ such that the distance between any two distinct points is $1$ when at least one of them is outside $X^star$ and such that, if $m,ninmathbb{Z}^+$ and $mneq n$, then$$d(x_m,x_n)=begin{cases}leftlvertfrac1m-frac1nrightrvert&text{ if }m,nneq0\frac1m&text{ if }n=0\frac1n&text{ if }m=0.end{cases}$$Then $x_0=lim_{ntoinfty}x_n$ and therefore this metric doesn't induce the discrete topology.

share|cite|improve this answer

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

$endgroup$

share|cite|improve this answer

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

answered 24 mins ago

José Carlos SantosJosé Carlos Santos

162k22130233

4

$begingroup$

Of course, take any nondiscrete metric space, say $Y$ and embed it into $X$ (this assumes $Y$ has cardinality lesser then $X$). This can be $mathbb{Q}$ or $mathbb{R}$ for example. You modify the metric a bit by using $d(x,y)=min(d_Y(x,y),1)$. So you now have nondiscrete metric for elements in $Ysubseteq X$. For any other pair define the distance as $1$ unless they are equal (which obviously has to be $0$).

share|cite|improve this answer

edited 19 mins ago

answered 24 mins ago

freakishfreakish

12.5k1630

$endgroup$

add a comment | 

4

$begingroup$

Of course, take any nondiscrete metric space, say $Y$ and embed it into $X$ (this assumes $Y$ has cardinality lesser then $X$). This can be $mathbb{Q}$ or $mathbb{R}$ for example. You modify the metric a bit by using $d(x,y)=min(d_Y(x,y),1)$. So you now have nondiscrete metric for elements in $Ysubseteq X$. For any other pair define the distance as $1$ unless they are equal (which obviously has to be $0$).

share|cite|improve this answer

edited 19 mins ago

answered 24 mins ago

freakishfreakish

12.5k1630

$endgroup$

add a comment | 

$begingroup$

Of course, take any nondiscrete metric space, say $Y$ and embed it into $X$ (this assumes $Y$ has cardinality lesser then $X$). This can be $mathbb{Q}$ or $mathbb{R}$ for example. You modify the metric a bit by using $d(x,y)=min(d_Y(x,y),1)$. So you now have nondiscrete metric for elements in $Ysubseteq X$. For any other pair define the distance as $1$ unless they are equal (which obviously has to be $0$).

share|cite|improve this answer

edited 19 mins ago

answered 24 mins ago

freakishfreakish

12.5k1630

$endgroup$

share|cite|improve this answer

edited 19 mins ago

answered 24 mins ago

freakishfreakish

12.5k1630

answered 24 mins ago

freakishfreakish

12.5k1630

answered 24 mins ago

freakishfreakish

12.5k1630

1

$begingroup$

If $X$ has cardinality $mathfrak{c}$, then the answer is yes: suppose that $phi : X to (0,1)$ is a bijection. Then define $d(x,y) = |phi(x) - phi(y)|$ for all $x,y in X$. This puts a topology on $X$ that is not discrete, since the usual topology on $Bbb{R}$ is not discrete.

If $X$ has cardinality greater than $mathfrak{c}$, then the answer is again yes because you can "cheat" by doing the following: choose a subset $U$ of $X$ of cardinality $mathfrak{c}$, and for all $x,y in U$ define (as before) $d(x,y) = |phi(x) - phi(y)|$. Put the discrete metric on $V = X setminus U$, that is, for all $x,y in X setminus U$, define $d(x,y) = 1$ if $xneq y$ and $d(x,y) =0$ if $x=y$. And finally, define $d(x,y) = 1$ if $x in U$, and $y in V$ (or $x in V$ and $y in U$).

Then, one can check that the triangle inequality still holds. The topology this metric induces is not discrete because of the way we have defined the metric on $U subset X$.

share|cite|improve this answer

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

$endgroup$

add a comment | 

1

$begingroup$

If $X$ has cardinality $mathfrak{c}$, then the answer is yes: suppose that $phi : X to (0,1)$ is a bijection. Then define $d(x,y) = |phi(x) - phi(y)|$ for all $x,y in X$. This puts a topology on $X$ that is not discrete, since the usual topology on $Bbb{R}$ is not discrete.

If $X$ has cardinality greater than $mathfrak{c}$, then the answer is again yes because you can "cheat" by doing the following: choose a subset $U$ of $X$ of cardinality $mathfrak{c}$, and for all $x,y in U$ define (as before) $d(x,y) = |phi(x) - phi(y)|$. Put the discrete metric on $V = X setminus U$, that is, for all $x,y in X setminus U$, define $d(x,y) = 1$ if $xneq y$ and $d(x,y) =0$ if $x=y$. And finally, define $d(x,y) = 1$ if $x in U$, and $y in V$ (or $x in V$ and $y in U$).

Then, one can check that the triangle inequality still holds. The topology this metric induces is not discrete because of the way we have defined the metric on $U subset X$.

share|cite|improve this answer

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

$endgroup$

add a comment | 

$begingroup$

If $X$ has cardinality $mathfrak{c}$, then the answer is yes: suppose that $phi : X to (0,1)$ is a bijection. Then define $d(x,y) = |phi(x) - phi(y)|$ for all $x,y in X$. This puts a topology on $X$ that is not discrete, since the usual topology on $Bbb{R}$ is not discrete.

If $X$ has cardinality greater than $mathfrak{c}$, then the answer is again yes because you can "cheat" by doing the following: choose a subset $U$ of $X$ of cardinality $mathfrak{c}$, and for all $x,y in U$ define (as before) $d(x,y) = |phi(x) - phi(y)|$. Put the discrete metric on $V = X setminus U$, that is, for all $x,y in X setminus U$, define $d(x,y) = 1$ if $xneq y$ and $d(x,y) =0$ if $x=y$. And finally, define $d(x,y) = 1$ if $x in U$, and $y in V$ (or $x in V$ and $y in U$).

Then, one can check that the triangle inequality still holds. The topology this metric induces is not discrete because of the way we have defined the metric on $U subset X$.

share|cite|improve this answer

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

$endgroup$

share|cite|improve this answer

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363

answered 20 mins ago

BrahadeeshBrahadeesh

6,37442363