A convergent sequence has precisely one accumulation point
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As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point
Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.
I wish to prove:
Lemma: A convergent sequence has precisely one accumulation point.
My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.
If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.
I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.
real-analysis sequences-and-series limits
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
$endgroup$
add a comment |
$begingroup$
As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point
Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.
I wish to prove:
Lemma: A convergent sequence has precisely one accumulation point.
My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.
If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.
I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.
real-analysis sequences-and-series limits
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
$endgroup$
add a comment |
$begingroup$
As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point
Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.
I wish to prove:
Lemma: A convergent sequence has precisely one accumulation point.
My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.
If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.
I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.
real-analysis sequences-and-series limits
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
$endgroup$
As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point
Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.
I wish to prove:
Lemma: A convergent sequence has precisely one accumulation point.
My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.
If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.
I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
asked Jan 8 at 14:05
Wesley StrikWesley Strik
2,194424
2,194424
add a comment |
add a comment |
2 Answers
2
active
oldest
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Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.
Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.
Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.
That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$
Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
add a comment |
$begingroup$
An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.
Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.
The proper definition is:
$x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.
Try to show your lemma now.
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.
Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.
Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.
That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$
Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
add a comment |
$begingroup$
Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.
Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.
Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.
That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$
Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
add a comment |
$begingroup$
Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.
Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.
Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.
That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$
Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.
Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.
Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.
That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$
Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 8 at 14:22
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
add a comment |
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
$begingroup$
Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 14:51
add a comment |
$begingroup$
An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.
Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.
The proper definition is:
$x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.
Try to show your lemma now.
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
add a comment |
$begingroup$
An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.
Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.
The proper definition is:
$x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.
Try to show your lemma now.
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
add a comment |
$begingroup$
An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.
Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.
The proper definition is:
$x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.
Try to show your lemma now.
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
$endgroup$
An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.
Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.
The proper definition is:
$x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.
Try to show your lemma now.
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
answered Jan 8 at 14:13
mechanodroidmechanodroid
29k62648
29k62648
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
add a comment |
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
$begingroup$
Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
$endgroup$
– Wesley Strik
Jan 8 at 19:59
add a comment |
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