Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$?

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q neq 0$ such that $qin E$.

A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $mathbbR$ which has an interior point which is not a limit point of $E$.

real-analysis metric-spaces

share|cite|improve this question

edited Jul 25 at 9:56

asked Jul 25 at 9:51

Madhu

699922

add a comment | 

up vote
0
down vote

favorite

According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q neq 0$ such that $qin E$.

A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $mathbbR$ which has an interior point which is not a limit point of $E$.

real-analysis metric-spaces

share|cite|improve this question

edited Jul 25 at 9:56

asked Jul 25 at 9:51

Madhu

699922

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q neq 0$ such that $qin E$.

A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $mathbbR$ which has an interior point which is not a limit point of $E$.

real-analysis metric-spaces

share|cite|improve this question

edited Jul 25 at 9:56

asked Jul 25 at 9:51

Madhu

699922

According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q neq 0$ such that $qin E$.

A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $mathbbR$ which has an interior point which is not a limit point of $E$.

real-analysis metric-spaces

share|cite|improve this question

edited Jul 25 at 9:56

asked Jul 25 at 9:51

Madhu

699922

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 9:56

edited Jul 25 at 9:56

edited Jul 25 at 9:56

asked Jul 25 at 9:51

Madhu

699922

asked Jul 25 at 9:51

Madhu

699922

asked Jul 25 at 9:51

Madhu

699922

699922

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

If you are asking about $mathbbR$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $XsubseteqmathbbR$ and that $xin X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $xin U$ and $Usubseteq X$. Since the topology on $mathbbR$ is generated by open intervals, we know that there is an open interval $I$ so that $xin I$, $Isubseteq Usubseteq X$. Observe that for $n$ sufficiently large, $x+frac1n$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $mathbbR$, i.e.,
$$
d(x,y)=begincases0&x=y\
1&xnot=y
endcases.
$$
Suppose that $XsubseteqmathbbR$ and that $xin X$. We observe that $x$ is an interior point because the open ball $Bleft(x,frac12right)=xsubseteq X$. On the other hand, $x$ is not a limit point because the open ball $Bleft(x,frac12right)$ is an open set containing $x$ and no other points of $X$.

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence for any metric space with a metric other than discrete metric interior points should be limit points.
  – Madhu
  Jul 25 at 11:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And without isolated points (in the chosen metric)
  – Michael Burr
  Jul 25 at 12:34
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862243%2fis-interior-points-of-a-subset-e-of-a-metric-space-x-is-always-a-limit-point%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

If you are asking about $mathbbR$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $XsubseteqmathbbR$ and that $xin X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $xin U$ and $Usubseteq X$. Since the topology on $mathbbR$ is generated by open intervals, we know that there is an open interval $I$ so that $xin I$, $Isubseteq Usubseteq X$. Observe that for $n$ sufficiently large, $x+frac1n$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $mathbbR$, i.e.,
$$
d(x,y)=begincases0&x=y\
1&xnot=y
endcases.
$$
Suppose that $XsubseteqmathbbR$ and that $xin X$. We observe that $x$ is an interior point because the open ball $Bleft(x,frac12right)=xsubseteq X$. On the other hand, $x$ is not a limit point because the open ball $Bleft(x,frac12right)$ is an open set containing $x$ and no other points of $X$.

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence for any metric space with a metric other than discrete metric interior points should be limit points.
  – Madhu
  Jul 25 at 11:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And without isolated points (in the chosen metric)
  – Michael Burr
  Jul 25 at 12:34
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

If you are asking about $mathbbR$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $XsubseteqmathbbR$ and that $xin X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $xin U$ and $Usubseteq X$. Since the topology on $mathbbR$ is generated by open intervals, we know that there is an open interval $I$ so that $xin I$, $Isubseteq Usubseteq X$. Observe that for $n$ sufficiently large, $x+frac1n$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $mathbbR$, i.e.,
$$
d(x,y)=begincases0&x=y\
1&xnot=y
endcases.
$$
Suppose that $XsubseteqmathbbR$ and that $xin X$. We observe that $x$ is an interior point because the open ball $Bleft(x,frac12right)=xsubseteq X$. On the other hand, $x$ is not a limit point because the open ball $Bleft(x,frac12right)$ is an open set containing $x$ and no other points of $X$.

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence for any metric space with a metric other than discrete metric interior points should be limit points.
  – Madhu
  Jul 25 at 11:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And without isolated points (in the chosen metric)
  – Michael Burr
  Jul 25 at 12:34
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

If you are asking about $mathbbR$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $XsubseteqmathbbR$ and that $xin X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $xin U$ and $Usubseteq X$. Since the topology on $mathbbR$ is generated by open intervals, we know that there is an open interval $I$ so that $xin I$, $Isubseteq Usubseteq X$. Observe that for $n$ sufficiently large, $x+frac1n$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $mathbbR$, i.e.,
$$
d(x,y)=begincases0&x=y\
1&xnot=y
endcases.
$$
Suppose that $XsubseteqmathbbR$ and that $xin X$. We observe that $x$ is an interior point because the open ball $Bleft(x,frac12right)=xsubseteq X$. On the other hand, $x$ is not a limit point because the open ball $Bleft(x,frac12right)$ is an open set containing $x$ and no other points of $X$.

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

If you are asking about $mathbbR$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $XsubseteqmathbbR$ and that $xin X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $xin U$ and $Usubseteq X$. Since the topology on $mathbbR$ is generated by open intervals, we know that there is an open interval $I$ so that $xin I$, $Isubseteq Usubseteq X$. Observe that for $n$ sufficiently large, $x+frac1n$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $mathbbR$, i.e.,
$$
d(x,y)=begincases0&x=y\
1&xnot=y
endcases.
$$
Suppose that $XsubseteqmathbbR$ and that $xin X$. We observe that $x$ is an interior point because the open ball $Bleft(x,frac12right)=xsubseteq X$. On the other hand, $x$ is not a limit point because the open ball $Bleft(x,frac12right)$ is an open set containing $x$ and no other points of $X$.

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 10:01

Michael Burr

25.6k13262

answered Jul 25 at 10:01

Michael Burr

25.6k13262

answered Jul 25 at 10:01

Michael Burr

25.6k13262

25.6k13262

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence for any metric space with a metric other than discrete metric interior points should be limit points.
  – Madhu
  Jul 25 at 11:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And without isolated points (in the chosen metric)
  – Michael Burr
  Jul 25 at 12:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence for any metric space with a metric other than discrete metric interior points should be limit points.
  – Madhu
  Jul 25 at 11:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And without isolated points (in the chosen metric)
  – Michael Burr
  Jul 25 at 12:34
  

  
  
  
  

Hence for any metric space with a metric other than discrete metric interior points should be limit points.
– Madhu
Jul 25 at 11:49

Hence for any metric space with a metric other than discrete metric interior points should be limit points.
– Madhu
Jul 25 at 11:49

And without isolated points (in the chosen metric)
– Michael Burr
Jul 25 at 12:34

And without isolated points (in the chosen metric)
– Michael Burr
Jul 25 at 12:34

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862243%2fis-interior-points-of-a-subset-e-of-a-metric-space-x-is-always-a-limit-point%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email