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Why does the definition of convergence have $<varepsilon$ instead of $levarepsilon$? [duplicate]
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Why does the definition of limits of a function have strict inequality?
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So the definition of convergence for a real sequence that I'm looking at says that $(a_n)to L$ means that $forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$.
The book I am reading says some people will say $n>N$ and some will say $nge N$, and it doesn't matter. I agree with that part. But what about the $<varepsilon$? The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
I thought maybe the problem would be that a lot of points would sit exactly at say, $L+varepsilon$, but they can't sit there forever since we would be able to make $varepsilon$ smaller and so they'd still have to get closer to $L$ anyway. Then I thought maybe the problem would be that we get an interval with only one point) like $[2,2]$, and then that would make it so convergence means the sequence's terms have to equal $L$, but since $varepsilon$ is positive, I don't think the one-element-interval scenario can happen.
Why is $<varepsilon$ used (or required) to accurately describe what convergence is saying about a sequence, instead of $levarepsilon$?
(If anyone has the time, I would greatly appreciate any explanation as to why convergence is something we want to know about a real sequence or why people want to know about limits of real sequences in general so I may be able to appreciate this definition. All I understand so far is that some infinitely long, ordered lists of real numbers get arbitrarily close to a real number and some don't. Sometimes when reading examples it does surprise me what number a sequence converges to or that a sequence doesn't actually get arbitrarily close to a number it seems to get really close to, but this is all the motivation I can think of.)
real-analysis limits convergence real-numbers
asked Jul 15 at 11:48
anonanon444
1396
marked as duplicate by Hans Lundmark, Community♦ Jul 15 at 12:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
2
down vote
favorite
This question already has an answer here:
Why does the definition of limits of a function have strict inequality?
3 answers
So the definition of convergence for a real sequence that I'm looking at says that $(a_n)to L$ means that $forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$.
The book I am reading says some people will say $n>N$ and some will say $nge N$, and it doesn't matter. I agree with that part. But what about the $<varepsilon$? The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
I thought maybe the problem would be that a lot of points would sit exactly at say, $L+varepsilon$, but they can't sit there forever since we would be able to make $varepsilon$ smaller and so they'd still have to get closer to $L$ anyway. Then I thought maybe the problem would be that we get an interval with only one point) like $[2,2]$, and then that would make it so convergence means the sequence's terms have to equal $L$, but since $varepsilon$ is positive, I don't think the one-element-interval scenario can happen.
Why is $<varepsilon$ used (or required) to accurately describe what convergence is saying about a sequence, instead of $levarepsilon$?
(If anyone has the time, I would greatly appreciate any explanation as to why convergence is something we want to know about a real sequence or why people want to know about limits of real sequences in general so I may be able to appreciate this definition. All I understand so far is that some infinitely long, ordered lists of real numbers get arbitrarily close to a real number and some don't. Sometimes when reading examples it does surprise me what number a sequence converges to or that a sequence doesn't actually get arbitrarily close to a number it seems to get really close to, but this is all the motivation I can think of.)
real-analysis limits convergence real-numbers
asked Jul 15 at 11:48
anonanon444
1396
marked as duplicate by Hans Lundmark, Community♦ Jul 15 at 12:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Why does the definition of limits of a function have strict inequality?
3 answers
So the definition of convergence for a real sequence that I'm looking at says that $(a_n)to L$ means that $forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$.
The book I am reading says some people will say $n>N$ and some will say $nge N$, and it doesn't matter. I agree with that part. But what about the $<varepsilon$? The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
I thought maybe the problem would be that a lot of points would sit exactly at say, $L+varepsilon$, but they can't sit there forever since we would be able to make $varepsilon$ smaller and so they'd still have to get closer to $L$ anyway. Then I thought maybe the problem would be that we get an interval with only one point) like $[2,2]$, and then that would make it so convergence means the sequence's terms have to equal $L$, but since $varepsilon$ is positive, I don't think the one-element-interval scenario can happen.
Why is $<varepsilon$ used (or required) to accurately describe what convergence is saying about a sequence, instead of $levarepsilon$?
(If anyone has the time, I would greatly appreciate any explanation as to why convergence is something we want to know about a real sequence or why people want to know about limits of real sequences in general so I may be able to appreciate this definition. All I understand so far is that some infinitely long, ordered lists of real numbers get arbitrarily close to a real number and some don't. Sometimes when reading examples it does surprise me what number a sequence converges to or that a sequence doesn't actually get arbitrarily close to a number it seems to get really close to, but this is all the motivation I can think of.)
real-analysis limits convergence real-numbers
asked Jul 15 at 11:48
anonanon444
1396
This question already has an answer here:
Why does the definition of limits of a function have strict inequality?
3 answers
So the definition of convergence for a real sequence that I'm looking at says that $(a_n)to L$ means that $forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$.
The book I am reading says some people will say $n>N$ and some will say $nge N$, and it doesn't matter. I agree with that part. But what about the $<varepsilon$? The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
I thought maybe the problem would be that a lot of points would sit exactly at say, $L+varepsilon$, but they can't sit there forever since we would be able to make $varepsilon$ smaller and so they'd still have to get closer to $L$ anyway. Then I thought maybe the problem would be that we get an interval with only one point) like $[2,2]$, and then that would make it so convergence means the sequence's terms have to equal $L$, but since $varepsilon$ is positive, I don't think the one-element-interval scenario can happen.
Why is $<varepsilon$ used (or required) to accurately describe what convergence is saying about a sequence, instead of $levarepsilon$?
(If anyone has the time, I would greatly appreciate any explanation as to why convergence is something we want to know about a real sequence or why people want to know about limits of real sequences in general so I may be able to appreciate this definition. All I understand so far is that some infinitely long, ordered lists of real numbers get arbitrarily close to a real number and some don't. Sometimes when reading examples it does surprise me what number a sequence converges to or that a sequence doesn't actually get arbitrarily close to a number it seems to get really close to, but this is all the motivation I can think of.)
This question already has an answer here:
Why does the definition of limits of a function have strict inequality?
3 answers
real-analysis limits convergence real-numbers
asked Jul 15 at 11:48
anonanon444
1396
asked Jul 15 at 11:48
anonanon444
1396
asked Jul 15 at 11:48
anonanon444
1396
asked Jul 15 at 11:48
anonanon444
1396
1396
marked as duplicate by Hans Lundmark, Community♦ Jul 15 at 12:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Hans Lundmark, Community♦ Jul 15 at 12:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07
add a comment |Â
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07
add a comment |Â
4 Answers
4
active
oldest
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up vote
2
down vote
accepted
Let $(a_n)_ninmathbb N$ be a sequence of real numbers and let $linmathbb R$. Then the assertions
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon$
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilon$
are equivalent. In fact, given $varepsilon>0$, it is clear that, if $Ninmathbb N$ is such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon,$$then$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilontag1$$will also hold. And if the second assertion holds and if you pick $Ninmathbb N$ such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantfracvarepsilon2,$$then $(1)$ will also hold, since $fracvarepsilon2<varepsilon$.
If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
add a comment |Â
up vote
2
down vote
The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)
However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
add a comment |Â
up vote
2
down vote
Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)
$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$
$forall,varepsilon '> 0,exists NinmathbbN:n>Nimplies|a_n-L|leqvarepsilon '$
answered Jul 15 at 11:56
Le Anh Dung
708318
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up vote
0
down vote
Note that $$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon iff forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|le varepsilon$$
The two expressions are logically equivalent due the universal quantifier.
Thus it really does not matter.
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $(a_n)_ninmathbb N$ be a sequence of real numbers and let $linmathbb R$. Then the assertions
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon$
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilon$
are equivalent. In fact, given $varepsilon>0$, it is clear that, if $Ninmathbb N$ is such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon,$$then$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilontag1$$will also hold. And if the second assertion holds and if you pick $Ninmathbb N$ such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantfracvarepsilon2,$$then $(1)$ will also hold, since $fracvarepsilon2<varepsilon$.
If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
add a comment |Â
up vote
2
down vote
accepted
Let $(a_n)_ninmathbb N$ be a sequence of real numbers and let $linmathbb R$. Then the assertions
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon$
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilon$
are equivalent. In fact, given $varepsilon>0$, it is clear that, if $Ninmathbb N$ is such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon,$$then$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilontag1$$will also hold. And if the second assertion holds and if you pick $Ninmathbb N$ such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantfracvarepsilon2,$$then $(1)$ will also hold, since $fracvarepsilon2<varepsilon$.
If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $(a_n)_ninmathbb N$ be a sequence of real numbers and let $linmathbb R$. Then the assertions
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon$
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilon$
are equivalent. In fact, given $varepsilon>0$, it is clear that, if $Ninmathbb N$ is such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon,$$then$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilontag1$$will also hold. And if the second assertion holds and if you pick $Ninmathbb N$ such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantfracvarepsilon2,$$then $(1)$ will also hold, since $fracvarepsilon2<varepsilon$.
If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
Let $(a_n)_ninmathbb N$ be a sequence of real numbers and let $linmathbb R$. Then the assertions
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon$
- $(forallvarepsilon>0)(exists NinmathbbN)(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilon$
are equivalent. In fact, given $varepsilon>0$, it is clear that, if $Ninmathbb N$ is such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|<varepsilon,$$then$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantvarepsilontag1$$will also hold. And if the second assertion holds and if you pick $Ninmathbb N$ such that$$(forall ninmathbbN):ngeqslant Nimplies|a_n-l|leqslantfracvarepsilon2,$$then $(1)$ will also hold, since $fracvarepsilon2<varepsilon$.
If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
answered Jul 15 at 11:57
José Carlos Santos
114k1698177
114k1698177
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
add a comment |Â
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
Some of us very much use the second one, thank you.
– Did
Jul 15 at 12:25
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
@Did I've never seen it in print. Are you aware of some example?
– José Carlos Santos
Jul 15 at 12:26
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
Some of us even explain to their students when they first encounter the definition of a limit that both inequalities, strict and large, work.
– Did
Jul 15 at 12:31
add a comment |Â
up vote
2
down vote
The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)
However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
add a comment |Â
up vote
2
down vote
The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)
However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)
However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
The book made no mention of that, and so I tried to figure out if $levarepsilon$ would work too. I feel like there must be a reason it has to be $<varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $nge N$.
All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)
However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
edited Jul 15 at 11:54
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
answered Jul 15 at 11:51
Stefan Mesken
13.6k32045
13.6k32045
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
add a comment |Â
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up!
– anonanon444
Jul 15 at 11:53
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
Well, wait, how come you think it's not meaningful?
– anonanon444
Jul 15 at 11:55
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
@anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics.
– Stefan Mesken
Jul 15 at 11:56
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify!
– anonanon444
Jul 15 at 11:58
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
@anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence.
– Stefan Mesken
Jul 15 at 11:59
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up vote
2
down vote
Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)
$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$
$forall,varepsilon '> 0,exists NinmathbbN:n>Nimplies|a_n-L|leqvarepsilon '$
answered Jul 15 at 11:56
Le Anh Dung
708318
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up vote
2
down vote
Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)
$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$
$forall,varepsilon '> 0,exists NinmathbbN:n>Nimplies|a_n-L|leqvarepsilon '$
answered Jul 15 at 11:56
Le Anh Dung
708318
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)
$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$
$forall,varepsilon '> 0,exists NinmathbbN:n>Nimplies|a_n-L|leqvarepsilon '$
answered Jul 15 at 11:56
Le Anh Dung
708318
Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)
$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon$
$forall,varepsilon '> 0,exists NinmathbbN:n>Nimplies|a_n-L|leqvarepsilon '$
answered Jul 15 at 11:56
Le Anh Dung
708318
answered Jul 15 at 11:56
Le Anh Dung
708318
answered Jul 15 at 11:56
Le Anh Dung
708318
answered Jul 15 at 11:56
Le Anh Dung
708318
708318
add a comment |Â
add a comment |Â
up vote
0
down vote
Note that $$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon iff forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|le varepsilon$$
The two expressions are logically equivalent due the universal quantifier.
Thus it really does not matter.
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
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up vote
0
down vote
Note that $$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon iff forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|le varepsilon$$
The two expressions are logically equivalent due the universal quantifier.
Thus it really does not matter.
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that $$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon iff forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|le varepsilon$$
The two expressions are logically equivalent due the universal quantifier.
Thus it really does not matter.
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
Note that $$forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|<varepsilon iff forall,varepsilon>0,exists NinmathbbN:n>Nimplies|a_n-L|le varepsilon$$
The two expressions are logically equivalent due the universal quantifier.
Thus it really does not matter.
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
answered Jul 15 at 11:56
Mohammad Riazi-Kermani
27.6k41852
27.6k41852
add a comment |Â
add a comment |Â
It also doesn't matter.
– Wojowu
Jul 15 at 11:49
Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $nge N$ but $<varepsilon$, so I thought the consistency in $<varepsilon$ must have meant something important.
– anonanon444
Jul 15 at 11:52
It's equivalent, and $<$ looks nicer than $le.$
– zhw.
Jul 15 at 15:07