Show $0$ is the only cluster point of the set $A = $$frac1n : ninmathbb N$
Clash Royale CLAN TAG#URR8PPP
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Need a check of working, I'm new to cluster points.
Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.
Does that suffice, or is there something to add?
real-analysis sequences-and-series
add a comment |Â
up vote
0
down vote
favorite
Need a check of working, I'm new to cluster points.
Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.
Does that suffice, or is there something to add?
real-analysis sequences-and-series
1
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
1
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
1
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
1
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Need a check of working, I'm new to cluster points.
Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.
Does that suffice, or is there something to add?
real-analysis sequences-and-series
Need a check of working, I'm new to cluster points.
Define $a_n = frac1n$ for all $ninmathbb N$. Clearly $A =$ $frac1n:ninmathbb N$ $=$ $(a_n)$. Now $a_n$ converges to $0$, thus every subsequence of $a_n$ also converges to $0$. Hence $0$ is the only cluster point of the set A.
Does that suffice, or is there something to add?
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Sep 10 at 3:40


Harambe
5,91121843
5,91121843
asked Sep 10 at 1:18
Florian Suess
329110
329110
1
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
1
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
1
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
1
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07
add a comment |Â
1
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
1
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
1
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
1
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07
1
1
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
1
1
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
1
1
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
1
1
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.
To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:
- $x_0 < 0$
- $x_0 > 1$
- $x_0 in (0,1]cap A$
- $x_0 in (0,1]setminus A$
I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.
Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.
add a comment |Â
up vote
1
down vote
If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,
Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
add a comment |Â
up vote
0
down vote
There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.
To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:
- $x_0 < 0$
- $x_0 > 1$
- $x_0 in (0,1]cap A$
- $x_0 in (0,1]setminus A$
I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.
Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.
add a comment |Â
up vote
2
down vote
accepted
Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.
To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:
- $x_0 < 0$
- $x_0 > 1$
- $x_0 in (0,1]cap A$
- $x_0 in (0,1]setminus A$
I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.
Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.
To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:
- $x_0 < 0$
- $x_0 > 1$
- $x_0 in (0,1]cap A$
- $x_0 in (0,1]setminus A$
I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.
Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.
Your argument shows that $0$ is a cluster point, since $(a_n)$ is a sequence in $Asetminus 0$ that converges to $A$. However, it doesn't prove that there are no other cluster points, since not all sequences in $Asetminus 0$ are subsequences of $(a_n)$.
To prove that no other cluster points exist, you can use the definition. That is, show that given any other $x_0 in mathbb R$, there exists $varepsilon > 0$ such that
$$[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing.$$
You can do this easily by splitting into cases:
- $x_0 < 0$
- $x_0 > 1$
- $x_0 in (0,1]cap A$
- $x_0 in (0,1]setminus A$
I'll prove one case for you, and by drawing pictures, you can figure out how to do the other three.
Suppose $x_0 > 1$. Let $varepsilon = fracx_0-12$. Then $[(x_0 - varepsilon, x_0 + varepsilon)cap A]setminus x_0 = varnothing$, so $A$ has no cluster points greater than $1$.
edited Sep 10 at 6:31
answered Sep 10 at 2:56


Harambe
5,91121843
5,91121843
add a comment |Â
add a comment |Â
up vote
1
down vote
If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,
Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
add a comment |Â
up vote
1
down vote
If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,
Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,
Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.
If this is for a class, I think that $0$ being the limit of the sequence should suffice as an argument for $0$ being a cluster point of $A$ (if you discussed this in class yet). However I would suggest being more explicit in your proof that this is the only cluster point of the set $A$. Something like,
Suppose that $p in mathbbR$ is a cluster point of the set $A$. Then $forall epsilon >0$, the set $a_n in A, $ contains an infinite number of points. By the least upper bound property of $mathbbN$, there is necessarily an index $i in mathbbN$ such that $a_i=min a_n in A, $. You can continue this argument to show that the subsequence starting with $a_1,a_2,...,a_i$, and continuing with the points of $a_n in A, $, ordered by their indices, is a subsequence of $a_n$ with limit $p$. Then by the uniqueness of the limit of a sequence, $p=0$.
answered Sep 10 at 1:48
clmundergrad
464
464
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
add a comment |Â
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Just in case you are not aware, you can edit your answer instead of posting a new one. Just click "edit" below the answer.
â user99914
Sep 10 at 3:50
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
Thanks, I just made my first stack exchange account, I didnt realize anyone could see a deleted answer
â clmundergrad
Sep 10 at 7:48
add a comment |Â
up vote
0
down vote
There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
add a comment |Â
up vote
0
down vote
There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.
There is indeed a flaw I've just realised. A subsequence of $a_n$ does not represent the entire set of subsequences in $A$. Hence I have not discovered what every subsequence in $A$ actually looks like.
edited Sep 10 at 1:53
answered Sep 10 at 1:46
Florian Suess
329110
329110
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
add a comment |Â
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
For example the constant sequence $frac12$ and so on. It is not a direct subsequence of $a_n$ hence I have not considered it.
â Florian Suess
Sep 10 at 1:47
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
I'm just hoping for confirmation from some professionals, because as I said, I'm new to cluster points.
â Florian Suess
Sep 10 at 1:48
add a comment |Â
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1
Whether or not your argument is correct depends on what you know about cluster points. What is the definition? Is the definition given in terms of sequences?
â user99914
Sep 10 at 1:30
1
The definition I've become familiar with; Let $A$ be a subset of $mathbb R$ and let $x_0 in mathbb R$, $x_0$ is a cluster point pf $A$ if for all $epsilon > 0$ the set $(x_0-epsilon,x_0+epsilon) cap Asetminus$$x_0$ $ne$ the empty set.
â Florian Suess
Sep 10 at 1:34
1
I've also come familiar with the synonymity of $x_0 $ is a cluster point AND there exists a sequence $(a_n)_ninmathbb N$ in $Asetminus$$x_0$ such that $lim_nrightarrowinftya_n = x_0$
â Florian Suess
Sep 10 at 1:36
1
Moreover I use the fact that if a sequence converges to say $L$, then every subsequence (of that sequence) must also converge to $L$
â Florian Suess
Sep 10 at 1:37
How are you defining "cluster point"? Typical definition: a point $x$ is a cluster point of $A$ if every open neighborhood of $x$ contains a point of $A$ other than $x$ itself, So, the fact that a sequence of points in $A$ converges to $x$ doesn't automatically make $x$ a cluster point (consider the constant sequence $a_n = x$).
â Bungo
Sep 10 at 3:07