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Find the set of all cluster points

How to find ALL cluster points of a sequence?Find the cluster points of a setFind all cluster points for the sequence $x_n$ = The $n$-th rational numberShowing set of all cluster points of sequence in extended $mathbb R $ is closed.Cluster points and the sequence 1,1,2,1,2,3,1,2,3,4,1,…Find interior points, boundary points, cluster points, limit points and isolated points of a setNon-countable set of points in a plane must have a cluster pointProve (or disprove) that $Bbb N$ is the set of all cluster points of $(x_n)$A set $S subset mathbbR$ has no cluster points iff $S cap [-n,n]$ is a finite set for each $n geq 1$Proving the set of all cluster points is closed

1

$begingroup$

Given the set $A=[a,b)$ with $a < b$, write the set of all cluster points of $A$.

Because it states $a < b$, I'm not sure how to start the question.

Are $a$ and $b$ still both cluster points?

For $b$ you can get $I=[b-e,b+e]$ with $e>0$ where we can find a point $y$ such that $b-e < y < b < b+e$ and $y>a$ then $y$ is an element of $A$ and $y$ doesn't equal $b$ so $b$ is a cluster point.

(I think the above is correct proof although I am new to cluster points)
But then for a, im not sure on how to prove it. thanks

real-analysis analysis limits

share|cite|improve this question

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

$endgroup$

bumped to the homepage by Community♦ yesterday

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $a$ and $b$ are both cluster points, but check your exact definition of cluster point, it may also include the points in the interior.
  $endgroup$
  – Gregory Grant
  May 14 '15 at 13:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you prove the interior are cluster points? thanks
  $endgroup$
  – Lauren Bathers
  May 14 '15 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The same way, any inverval $(c-epsilon,c+epsilon)$ will intersect $A-c$. But the definition varies from author to author. What is the exact definition of cluster point in your book?
  $endgroup$
  – Gregory Grant
  May 14 '15 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it says "Given a set A and a point x, we say that x is a cluster point for the set A if in any open interval I containing x there are infinitely many points of A. Intervals are in the form I=(x-ϵ,x+ϵ) for ϵ>0"
  $endgroup$
  – Lauren Bathers
  May 14 '15 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it, so in that case the interior points are also cluster points. It should be pretty clear why. You just need to show if $cin[a,b)$ then $(c-varepsilon,c+varepsilon)cap[a,b)$ has infinitely many points.
  $endgroup$
  – Gregory Grant
  May 14 '15 at 14:49
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Given the set $A=[a,b)$ with $a < b$, write the set of all cluster points of $A$.

Because it states $a < b$, I'm not sure how to start the question.

Are $a$ and $b$ still both cluster points?

For $b$ you can get $I=[b-e,b+e]$ with $e>0$ where we can find a point $y$ such that $b-e < y < b < b+e$ and $y>a$ then $y$ is an element of $A$ and $y$ doesn't equal $b$ so $b$ is a cluster point.

(I think the above is correct proof although I am new to cluster points)
But then for a, im not sure on how to prove it. thanks

real-analysis analysis limits

share|cite|improve this question

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

$endgroup$

bumped to the homepage by Community♦ yesterday

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $a$ and $b$ are both cluster points, but check your exact definition of cluster point, it may also include the points in the interior.
  $endgroup$
  – Gregory Grant
  May 14 '15 at 13:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you prove the interior are cluster points? thanks
  $endgroup$
  – Lauren Bathers
  May 14 '15 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The same way, any inverval $(c-epsilon,c+epsilon)$ will intersect $A-c$. But the definition varies from author to author. What is the exact definition of cluster point in your book?
  $endgroup$
  – Gregory Grant
  May 14 '15 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it says "Given a set A and a point x, we say that x is a cluster point for the set A if in any open interval I containing x there are infinitely many points of A. Intervals are in the form I=(x-ϵ,x+ϵ) for ϵ>0"
  $endgroup$
  – Lauren Bathers
  May 14 '15 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it, so in that case the interior points are also cluster points. It should be pretty clear why. You just need to show if $cin[a,b)$ then $(c-varepsilon,c+varepsilon)cap[a,b)$ has infinitely many points.
  $endgroup$
  – Gregory Grant
  May 14 '15 at 14:49
  
  

  
  
  
  

add a comment | 

$begingroup$

Given the set $A=[a,b)$ with $a < b$, write the set of all cluster points of $A$.

Because it states $a < b$, I'm not sure how to start the question.

Are $a$ and $b$ still both cluster points?

For $b$ you can get $I=[b-e,b+e]$ with $e>0$ where we can find a point $y$ such that $b-e < y < b < b+e$ and $y>a$ then $y$ is an element of $A$ and $y$ doesn't equal $b$ so $b$ is a cluster point.

(I think the above is correct proof although I am new to cluster points)
But then for a, im not sure on how to prove it. thanks

real-analysis analysis limits

share|cite|improve this question

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

$endgroup$

share|cite|improve this question

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

share|cite|improve this question

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

edited May 14 '15 at 13:53

mrf

37.6k64786

edited May 14 '15 at 13:53

mrf

37.6k64786

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

asked May 14 '15 at 13:08

Lauren BathersLauren Bathers

212112

1 Answer
1

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oldest

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0

$begingroup$

Given $epsilon>0$ (and $epsilon < b-a$), just take the mid-point between $a$ and $a+epsilon$. This proves that $a$ is a cluster point of $A$. Now you should prove, with similar arguments, that every interior point of $A$ is also a cluster point, so that the closure of $A$ is $[a,b]$.

Of course $a<b$, otherwise $A=emptyset$.

share|cite|improve this answer

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

$endgroup$

add a comment | 

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0

$begingroup$

Given $epsilon>0$ (and $epsilon < b-a$), just take the mid-point between $a$ and $a+epsilon$. This proves that $a$ is a cluster point of $A$. Now you should prove, with similar arguments, that every interior point of $A$ is also a cluster point, so that the closure of $A$ is $[a,b]$.

Of course $a<b$, otherwise $A=emptyset$.

share|cite|improve this answer

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

$endgroup$

add a comment | 

0

$begingroup$

Given $epsilon>0$ (and $epsilon < b-a$), just take the mid-point between $a$ and $a+epsilon$. This proves that $a$ is a cluster point of $A$. Now you should prove, with similar arguments, that every interior point of $A$ is also a cluster point, so that the closure of $A$ is $[a,b]$.

Of course $a<b$, otherwise $A=emptyset$.

share|cite|improve this answer

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

$endgroup$

add a comment | 

$begingroup$

Given $epsilon>0$ (and $epsilon < b-a$), just take the mid-point between $a$ and $a+epsilon$. This proves that $a$ is a cluster point of $A$. Now you should prove, with similar arguments, that every interior point of $A$ is also a cluster point, so that the closure of $A$ is $[a,b]$.

Of course $a<b$, otherwise $A=emptyset$.

share|cite|improve this answer

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

$endgroup$

share|cite|improve this answer

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469

answered May 14 '15 at 13:11

SiminoreSiminore

30.5k33469