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Isolated points in a complete metric space

0

$begingroup$

I need to prove that if $X$ is a countable (infinite) complete metric space, then $X$ has infinitely many isolated points.

I have read that the Category Baire theorem implies that X should have at least one isolated point, but I have no idea how to show the set of isolated points is infinite.

Thanks for any help!

general-topology metric-spaces baire-category

share|cite|improve this question

asked Apr 2 at 18:01

MathloverMathlover

898

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Remove an isolated point from a complete metric space. Does it remain complete? Now, think about induction.
  $endgroup$
  – Moishe Kohan
  Apr 2 at 18:04
  

  
  
  
  

add a comment | 

0

$begingroup$

I need to prove that if $X$ is a countable (infinite) complete metric space, then $X$ has infinitely many isolated points.

I have read that the Category Baire theorem implies that X should have at least one isolated point, but I have no idea how to show the set of isolated points is infinite.

Thanks for any help!

general-topology metric-spaces baire-category

share|cite|improve this question

asked Apr 2 at 18:01

MathloverMathlover

898

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Remove an isolated point from a complete metric space. Does it remain complete? Now, think about induction.
  $endgroup$
  – Moishe Kohan
  Apr 2 at 18:04
  

  
  
  
  

add a comment | 

$begingroup$

I need to prove that if $X$ is a countable (infinite) complete metric space, then $X$ has infinitely many isolated points.

I have read that the Category Baire theorem implies that X should have at least one isolated point, but I have no idea how to show the set of isolated points is infinite.

Thanks for any help!

general-topology metric-spaces baire-category

share|cite|improve this question

asked Apr 2 at 18:01

MathloverMathlover

898

$endgroup$

share|cite|improve this question

asked Apr 2 at 18:01

MathloverMathlover

898

share|cite|improve this question

asked Apr 2 at 18:01

MathloverMathlover

898

asked Apr 2 at 18:01

MathloverMathlover

898

asked Apr 2 at 18:01

MathloverMathlover

898

2 Answers
2

active

oldest

votes

1

$begingroup$

Suppose that it has a finite number of isolated points $x_1,..,x_n$, $Y=X-x_1,..,x_n$ is still complete. Every element $yin Y$ as an empty interior. You can write $Y=cup_ninmathbbNy_n$, Baire implies that $cup_ninmathbbNy_n=Y$ has an empty interior. Contradiction.

share|cite|improve this answer

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

$endgroup$

add a comment | 

0

$begingroup$

Well, as Moishe points out in the comments above, if $X$ has only finitely-many isolated points, then we can remove them all, and obtain a complete, countably-infinite metric space (say $Y$) with no isolated points.

But this is impossible! Let $f:Bbb Nto Y$ be a bijection, and consider the sets $O_n:=Ysetminusbiglf(n)bigr.$ Each $O_n$ is open (obviously) and dense (since $f(n)$ is not isolated in $Y$), so since $Y$ is a complete metric space, then BCT tells us that $bigcap_ninBbb NO_n=emptyset$ is dense in $Y,$ which is absurd.

share|cite|improve this answer

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

$endgroup$

add a comment | 

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1

$begingroup$

Suppose that it has a finite number of isolated points $x_1,..,x_n$, $Y=X-x_1,..,x_n$ is still complete. Every element $yin Y$ as an empty interior. You can write $Y=cup_ninmathbbNy_n$, Baire implies that $cup_ninmathbbNy_n=Y$ has an empty interior. Contradiction.

share|cite|improve this answer

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

$endgroup$

add a comment | 

1

$begingroup$

Suppose that it has a finite number of isolated points $x_1,..,x_n$, $Y=X-x_1,..,x_n$ is still complete. Every element $yin Y$ as an empty interior. You can write $Y=cup_ninmathbbNy_n$, Baire implies that $cup_ninmathbbNy_n=Y$ has an empty interior. Contradiction.

share|cite|improve this answer

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

$endgroup$

add a comment | 

$begingroup$

Suppose that it has a finite number of isolated points $x_1,..,x_n$, $Y=X-x_1,..,x_n$ is still complete. Every element $yin Y$ as an empty interior. You can write $Y=cup_ninmathbbNy_n$, Baire implies that $cup_ninmathbbNy_n=Y$ has an empty interior. Contradiction.

share|cite|improve this answer

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

$endgroup$

share|cite|improve this answer

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

answered Apr 2 at 18:19

Tsemo AristideTsemo Aristide

61k11446

0

$begingroup$

Well, as Moishe points out in the comments above, if $X$ has only finitely-many isolated points, then we can remove them all, and obtain a complete, countably-infinite metric space (say $Y$) with no isolated points.

But this is impossible! Let $f:Bbb Nto Y$ be a bijection, and consider the sets $O_n:=Ysetminusbiglf(n)bigr.$ Each $O_n$ is open (obviously) and dense (since $f(n)$ is not isolated in $Y$), so since $Y$ is a complete metric space, then BCT tells us that $bigcap_ninBbb NO_n=emptyset$ is dense in $Y,$ which is absurd.

share|cite|improve this answer

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

$endgroup$

add a comment | 

0

$begingroup$

Well, as Moishe points out in the comments above, if $X$ has only finitely-many isolated points, then we can remove them all, and obtain a complete, countably-infinite metric space (say $Y$) with no isolated points.

But this is impossible! Let $f:Bbb Nto Y$ be a bijection, and consider the sets $O_n:=Ysetminusbiglf(n)bigr.$ Each $O_n$ is open (obviously) and dense (since $f(n)$ is not isolated in $Y$), so since $Y$ is a complete metric space, then BCT tells us that $bigcap_ninBbb NO_n=emptyset$ is dense in $Y,$ which is absurd.

share|cite|improve this answer

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

$endgroup$

add a comment | 

$begingroup$

Well, as Moishe points out in the comments above, if $X$ has only finitely-many isolated points, then we can remove them all, and obtain a complete, countably-infinite metric space (say $Y$) with no isolated points.

But this is impossible! Let $f:Bbb Nto Y$ be a bijection, and consider the sets $O_n:=Ysetminusbiglf(n)bigr.$ Each $O_n$ is open (obviously) and dense (since $f(n)$ is not isolated in $Y$), so since $Y$ is a complete metric space, then BCT tells us that $bigcap_ninBbb NO_n=emptyset$ is dense in $Y,$ which is absurd.

share|cite|improve this answer

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

$endgroup$

share|cite|improve this answer

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162

answered Apr 2 at 18:23

Cameron BuieCameron Buie

87.4k773162