Mixing
Why is this definition of convergence incorrect?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
From my understanding, a sequence $a_n$ is said to converge to $a$ if for all $epsilon > 0$, there exists an index $N$ such that for all $n geq N$, we have
$$|a_n - a| < epsilon.$$
But, why is this different from the following definition:
There exists an index $N$ for every $epsilon > 0$, such that for all $n geq N$, we have
$$|a_n - a| < epsilon $$
Pretty much, I switched the $forall epsilon > 0$ quantifier with the $exists N$ quantifier, and the definition becomes invalid, but why?
real-analysis convergence
asked 3 hours ago
Hat
858115
add a comment |Â
up vote
3
down vote
favorite
From my understanding, a sequence $a_n$ is said to converge to $a$ if for all $epsilon > 0$, there exists an index $N$ such that for all $n geq N$, we have
$$|a_n - a| < epsilon.$$
But, why is this different from the following definition:
There exists an index $N$ for every $epsilon > 0$, such that for all $n geq N$, we have
$$|a_n - a| < epsilon $$
Pretty much, I switched the $forall epsilon > 0$ quantifier with the $exists N$ quantifier, and the definition becomes invalid, but why?
real-analysis convergence
asked 3 hours ago
Hat
858115
3
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
From my understanding, a sequence $a_n$ is said to converge to $a$ if for all $epsilon > 0$, there exists an index $N$ such that for all $n geq N$, we have
$$|a_n - a| < epsilon.$$
But, why is this different from the following definition:
There exists an index $N$ for every $epsilon > 0$, such that for all $n geq N$, we have
$$|a_n - a| < epsilon $$
Pretty much, I switched the $forall epsilon > 0$ quantifier with the $exists N$ quantifier, and the definition becomes invalid, but why?
real-analysis convergence
asked 3 hours ago
Hat
858115
From my understanding, a sequence $a_n$ is said to converge to $a$ if for all $epsilon > 0$, there exists an index $N$ such that for all $n geq N$, we have
$$|a_n - a| < epsilon.$$
But, why is this different from the following definition:
There exists an index $N$ for every $epsilon > 0$, such that for all $n geq N$, we have
$$|a_n - a| < epsilon $$
Pretty much, I switched the $forall epsilon > 0$ quantifier with the $exists N$ quantifier, and the definition becomes invalid, but why?
real-analysis convergence
real-analysis convergence
asked 3 hours ago
Hat
858115
asked 3 hours ago
Hat
858115
asked 3 hours ago
Hat
858115
asked 3 hours ago
Hat
858115
asked 3 hours ago
Hat
858115
858115
3
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago
add a comment |Â
3
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago
3
3
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Look at your second definition and take a $N$ such that for every $varepsilon > 0$, $forall n geq N$, $|a_n-a| leq varepsilon$.
This means that for all $n geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to
$$exists N in mathbbN, forall ngeq N, a_n = a$$
i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.
answered 2 hours ago
TheSilverDoe
1,20011
add a comment |Â
up vote
2
down vote
I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.
There is a difference between
$exists N$ such that $forall epsilon>0$, such that $forall n>N$, $|a_n - a| < epsilon$
and
$exists N_epsilon$ for all $epsilon>0$, such that $forall n>N_epsilon$, $|a_n - a| < epsilon$
You wrote the first (it is at least implied), hence the answers you got.
I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.
answered 2 hours ago
Martigan
4,899916
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Look at your second definition and take a $N$ such that for every $varepsilon > 0$, $forall n geq N$, $|a_n-a| leq varepsilon$.
This means that for all $n geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to
$$exists N in mathbbN, forall ngeq N, a_n = a$$
i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.
answered 2 hours ago
TheSilverDoe
1,20011
add a comment |Â
up vote
3
down vote
Look at your second definition and take a $N$ such that for every $varepsilon > 0$, $forall n geq N$, $|a_n-a| leq varepsilon$.
This means that for all $n geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to
$$exists N in mathbbN, forall ngeq N, a_n = a$$
i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.
answered 2 hours ago
TheSilverDoe
1,20011
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Look at your second definition and take a $N$ such that for every $varepsilon > 0$, $forall n geq N$, $|a_n-a| leq varepsilon$.
This means that for all $n geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to
$$exists N in mathbbN, forall ngeq N, a_n = a$$
i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.
answered 2 hours ago
TheSilverDoe
1,20011
Look at your second definition and take a $N$ such that for every $varepsilon > 0$, $forall n geq N$, $|a_n-a| leq varepsilon$.
This means that for all $n geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to
$$exists N in mathbbN, forall ngeq N, a_n = a$$
i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.
answered 2 hours ago
TheSilverDoe
1,20011
answered 2 hours ago
TheSilverDoe
1,20011
answered 2 hours ago
TheSilverDoe
1,20011
answered 2 hours ago
TheSilverDoe
1,20011
1,20011
add a comment |Â
add a comment |Â
up vote
2
down vote
I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.
There is a difference between
$exists N$ such that $forall epsilon>0$, such that $forall n>N$, $|a_n - a| < epsilon$
and
$exists N_epsilon$ for all $epsilon>0$, such that $forall n>N_epsilon$, $|a_n - a| < epsilon$
You wrote the first (it is at least implied), hence the answers you got.
I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.
answered 2 hours ago
Martigan
4,899916
add a comment |Â
up vote
2
down vote
I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.
There is a difference between
$exists N$ such that $forall epsilon>0$, such that $forall n>N$, $|a_n - a| < epsilon$
and
$exists N_epsilon$ for all $epsilon>0$, such that $forall n>N_epsilon$, $|a_n - a| < epsilon$
You wrote the first (it is at least implied), hence the answers you got.
I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.
answered 2 hours ago
Martigan
4,899916
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.
There is a difference between
$exists N$ such that $forall epsilon>0$, such that $forall n>N$, $|a_n - a| < epsilon$
and
$exists N_epsilon$ for all $epsilon>0$, such that $forall n>N_epsilon$, $|a_n - a| < epsilon$
You wrote the first (it is at least implied), hence the answers you got.
I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.
answered 2 hours ago
Martigan
4,899916
I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.
There is a difference between
$exists N$ such that $forall epsilon>0$, such that $forall n>N$, $|a_n - a| < epsilon$
and
$exists N_epsilon$ for all $epsilon>0$, such that $forall n>N_epsilon$, $|a_n - a| < epsilon$
You wrote the first (it is at least implied), hence the answers you got.
I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.
answered 2 hours ago
Martigan
4,899916
answered 2 hours ago
Martigan
4,899916
answered 2 hours ago
Martigan
4,899916
answered 2 hours ago
Martigan
4,899916
4,899916
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2937799%2fwhy-is-this-definition-of-convergence-incorrect%23new-answer', 'question_page');
);
Post as a guest
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
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
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
3
think of $lbrace 1/n rbrace $, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $mid 1/n mid < epsilon$ $forall epsilon>0$, $forall n>N$ which is simply not possible.
– Baol
2 hours ago