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Is there a minimum of points that need to be $epsilon$-close to a point to call this point a limit point?
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I was thinking about the definition of a limit point and couldn't find a definition that contained any information if there is a minimum amount of points that need to be $epsilon$-close to a point $x_0$ to call this point a limit point.
So I tried to come up with a sequence that could look something like that for example:
$$a_n:=left{beginarrayll frac1n, & n not in [m-1,m] \
x_0, & n= m \
x_0-epsilon, & n=m-1endarrayright. qquad n,m in mathbbN$$
(First of all I am not sure if this is even a legitimate definition of a sequence, especially the $epsilon$ in the defintion, not to mention the correctness of the notation.)
If a sequence like this exists then this would be a null-sequence outside of the interval $[m-1,m]$ but would have a limit point at $x_0$.
Is this a correct statement or did I miss some details of the definition?
The definition I used was:
A number $p in mathbbR$ is a limit point of a set $M$, if for
every $epsilon > 0$ an element $k in M$ exists such that $k neq p$
and $|k-p| < epsilon$
Alternatively I also thought about defining a $xi>0$ which is infinitely small and then write my sequence as $a_n:=left{beginarrayll ... \
x_0-xi, & n=m-1endarrayright .$, so that I can get rid of the $epsilon$ in my sequence definition. I just wanted to see if the usage of $epsilon$ inside of a sequence is even allowed
real-analysis analysis limits definition
asked Jul 21 at 18:00
MLK
66112
add a comment |Â
up vote
0
down vote
favorite
I was thinking about the definition of a limit point and couldn't find a definition that contained any information if there is a minimum amount of points that need to be $epsilon$-close to a point $x_0$ to call this point a limit point.
So I tried to come up with a sequence that could look something like that for example:
$$a_n:=left{beginarrayll frac1n, & n not in [m-1,m] \
x_0, & n= m \
x_0-epsilon, & n=m-1endarrayright. qquad n,m in mathbbN$$
(First of all I am not sure if this is even a legitimate definition of a sequence, especially the $epsilon$ in the defintion, not to mention the correctness of the notation.)
If a sequence like this exists then this would be a null-sequence outside of the interval $[m-1,m]$ but would have a limit point at $x_0$.
Is this a correct statement or did I miss some details of the definition?
The definition I used was:
A number $p in mathbbR$ is a limit point of a set $M$, if for
every $epsilon > 0$ an element $k in M$ exists such that $k neq p$
and $|k-p| < epsilon$
Alternatively I also thought about defining a $xi>0$ which is infinitely small and then write my sequence as $a_n:=left{beginarrayll ... \
x_0-xi, & n=m-1endarrayright .$, so that I can get rid of the $epsilon$ in my sequence definition. I just wanted to see if the usage of $epsilon$ inside of a sequence is even allowed
real-analysis analysis limits definition
asked Jul 21 at 18:00
MLK
66112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was thinking about the definition of a limit point and couldn't find a definition that contained any information if there is a minimum amount of points that need to be $epsilon$-close to a point $x_0$ to call this point a limit point.
So I tried to come up with a sequence that could look something like that for example:
$$a_n:=left{beginarrayll frac1n, & n not in [m-1,m] \
x_0, & n= m \
x_0-epsilon, & n=m-1endarrayright. qquad n,m in mathbbN$$
(First of all I am not sure if this is even a legitimate definition of a sequence, especially the $epsilon$ in the defintion, not to mention the correctness of the notation.)
If a sequence like this exists then this would be a null-sequence outside of the interval $[m-1,m]$ but would have a limit point at $x_0$.
Is this a correct statement or did I miss some details of the definition?
The definition I used was:
A number $p in mathbbR$ is a limit point of a set $M$, if for
every $epsilon > 0$ an element $k in M$ exists such that $k neq p$
and $|k-p| < epsilon$
Alternatively I also thought about defining a $xi>0$ which is infinitely small and then write my sequence as $a_n:=left{beginarrayll ... \
x_0-xi, & n=m-1endarrayright .$, so that I can get rid of the $epsilon$ in my sequence definition. I just wanted to see if the usage of $epsilon$ inside of a sequence is even allowed
real-analysis analysis limits definition
asked Jul 21 at 18:00
MLK
66112
I was thinking about the definition of a limit point and couldn't find a definition that contained any information if there is a minimum amount of points that need to be $epsilon$-close to a point $x_0$ to call this point a limit point.
So I tried to come up with a sequence that could look something like that for example:
$$a_n:=left{beginarrayll frac1n, & n not in [m-1,m] \
x_0, & n= m \
x_0-epsilon, & n=m-1endarrayright. qquad n,m in mathbbN$$
(First of all I am not sure if this is even a legitimate definition of a sequence, especially the $epsilon$ in the defintion, not to mention the correctness of the notation.)
If a sequence like this exists then this would be a null-sequence outside of the interval $[m-1,m]$ but would have a limit point at $x_0$.
Is this a correct statement or did I miss some details of the definition?
The definition I used was:
A number $p in mathbbR$ is a limit point of a set $M$, if for
every $epsilon > 0$ an element $k in M$ exists such that $k neq p$
and $|k-p| < epsilon$
Alternatively I also thought about defining a $xi>0$ which is infinitely small and then write my sequence as $a_n:=left{beginarrayll ... \
x_0-xi, & n=m-1endarrayright .$, so that I can get rid of the $epsilon$ in my sequence definition. I just wanted to see if the usage of $epsilon$ inside of a sequence is even allowed
real-analysis analysis limits definition
asked Jul 21 at 18:00
MLK
66112
asked Jul 21 at 18:00
MLK
66112
asked Jul 21 at 18:00
MLK
66112
asked Jul 21 at 18:00
MLK
66112
66112
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The issue is that your sequence depends on $epsilon$, and therefore isn't a limit point. Here's a fact:
If $x_0$ is a limit point of some set $S$ and $x_0 notin S$, then for each $epsilon > 0$ there are infinitely many elements of $S$ within $epsilon$ of $x_0$.
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $epsilon$ so that only finitely many members of $S$ are within $epsilon$ of $x_0$. Let's enumerate them, $s_1,ldots,s_n$. We may then set $delta = min_1 leq j leq n |x_0 - s_j|$ and note that $S$ has no elements within distance $delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.
answered Jul 21 at 18:23
Marcus M
8,1731847
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
add a comment |Â
up vote
1
down vote
All you need is one for each $epsilon$.
answered Jul 21 at 18:04
marty cohen
69.2k446122
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The issue is that your sequence depends on $epsilon$, and therefore isn't a limit point. Here's a fact:
If $x_0$ is a limit point of some set $S$ and $x_0 notin S$, then for each $epsilon > 0$ there are infinitely many elements of $S$ within $epsilon$ of $x_0$.
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $epsilon$ so that only finitely many members of $S$ are within $epsilon$ of $x_0$. Let's enumerate them, $s_1,ldots,s_n$. We may then set $delta = min_1 leq j leq n |x_0 - s_j|$ and note that $S$ has no elements within distance $delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.
answered Jul 21 at 18:23
Marcus M
8,1731847
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
add a comment |Â
up vote
2
down vote
accepted
The issue is that your sequence depends on $epsilon$, and therefore isn't a limit point. Here's a fact:
If $x_0$ is a limit point of some set $S$ and $x_0 notin S$, then for each $epsilon > 0$ there are infinitely many elements of $S$ within $epsilon$ of $x_0$.
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $epsilon$ so that only finitely many members of $S$ are within $epsilon$ of $x_0$. Let's enumerate them, $s_1,ldots,s_n$. We may then set $delta = min_1 leq j leq n |x_0 - s_j|$ and note that $S$ has no elements within distance $delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.
answered Jul 21 at 18:23
Marcus M
8,1731847
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The issue is that your sequence depends on $epsilon$, and therefore isn't a limit point. Here's a fact:
If $x_0$ is a limit point of some set $S$ and $x_0 notin S$, then for each $epsilon > 0$ there are infinitely many elements of $S$ within $epsilon$ of $x_0$.
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $epsilon$ so that only finitely many members of $S$ are within $epsilon$ of $x_0$. Let's enumerate them, $s_1,ldots,s_n$. We may then set $delta = min_1 leq j leq n |x_0 - s_j|$ and note that $S$ has no elements within distance $delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.
answered Jul 21 at 18:23
Marcus M
8,1731847
The issue is that your sequence depends on $epsilon$, and therefore isn't a limit point. Here's a fact:
If $x_0$ is a limit point of some set $S$ and $x_0 notin S$, then for each $epsilon > 0$ there are infinitely many elements of $S$ within $epsilon$ of $x_0$.
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $epsilon$ so that only finitely many members of $S$ are within $epsilon$ of $x_0$. Let's enumerate them, $s_1,ldots,s_n$. We may then set $delta = min_1 leq j leq n |x_0 - s_j|$ and note that $S$ has no elements within distance $delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.
answered Jul 21 at 18:23
Marcus M
8,1731847
answered Jul 21 at 18:23
Marcus M
8,1731847
answered Jul 21 at 18:23
Marcus M
8,1731847
answered Jul 21 at 18:23
Marcus M
8,1731847
8,1731847
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
add a comment |Â
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
Wonderful. Thank you for answering my question. However I still can't fully comprehend why the proof by contrapositive has to imply infinitely many members of $S$ and not just one member of $S$ that is infinitely close to a point $x_0$?
– MLK
Jul 21 at 18:36
1
1
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
@MLK, that's because in the real numbers (and more generally any metric space) there's no notion of "infinitely close." If $x neq y$ are two points, then $|x - y| > 0$ is a number, and there is always a smaller positive number than that distance.
– Marcus M
Jul 21 at 18:58
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
Ah well. So I just defined something that is not allowed to exist. Thank you very much for the clarification.
– MLK
Jul 21 at 19:04
add a comment |Â
up vote
1
down vote
All you need is one for each $epsilon$.
answered Jul 21 at 18:04
marty cohen
69.2k446122
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
add a comment |Â
up vote
1
down vote
All you need is one for each $epsilon$.
answered Jul 21 at 18:04
marty cohen
69.2k446122
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
All you need is one for each $epsilon$.
answered Jul 21 at 18:04
marty cohen
69.2k446122
All you need is one for each $epsilon$.
answered Jul 21 at 18:04
marty cohen
69.2k446122
answered Jul 21 at 18:04
marty cohen
69.2k446122
answered Jul 21 at 18:04
marty cohen
69.2k446122
answered Jul 21 at 18:04
marty cohen
69.2k446122
69.2k446122
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
add a comment |Â
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
But theoretically it could be the same value $x$ for every $epsilon$?
– MLK
Jul 21 at 18:06
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@MLK, can $xne y$ and $x$ be $epsilon$-close to $y$ for every $epsilon >0$? (In a metric space, I assume?)
– jgon
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
@jgon yes they can in this case
– MLK
Jul 21 at 18:20
add a comment |Â
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