Is showing that $x_n rightarrow x_0Rightarrow f(x_n) rightarrow f(x_0)$ for a single sequence enough to prove...
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For all my homework in real analysis, when I've been asked to show that a function is continuous, I just found a single $x_n in D$ and showed that when $x_n rightarrow x_0$, $f(x_n) rightarrow f(x_0)$. Apparently, the sequence definition (as opposed to the epsilon delta definition) is (basically) only used to prove a function is not continuous, and I can't prove a function is continuous because then I'd have to show this is true for all possible sequences? Am I doing the math wrongly? Should I always use the epsilon delta definition when trying to prove that a function is continuous?
real-analysis continuity
edited May 29 at 9:53
glS
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asked May 27 at 0:45
MinYoung KimMinYoung Kim
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show 14 more comments
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For all my homework in real analysis, when I've been asked to show that a function is continuous, I just found a single $x_n in D$ and showed that when $x_n rightarrow x_0$, $f(x_n) rightarrow f(x_0)$. Apparently, the sequence definition (as opposed to the epsilon delta definition) is (basically) only used to prove a function is not continuous, and I can't prove a function is continuous because then I'd have to show this is true for all possible sequences? Am I doing the math wrongly? Should I always use the epsilon delta definition when trying to prove that a function is continuous?
real-analysis continuity
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
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1
$begingroup$
That would be incorrect.
$endgroup$
– Math1000
May 27 at 0:53
22
$begingroup$
oh god. this is not good.
$endgroup$
– MinYoung Kim
May 27 at 0:54
4
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
$endgroup$
– Arturo Magidin
May 27 at 0:55
27
$begingroup$
Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
$endgroup$
– Arturo Magidin
May 27 at 0:55
3
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
$endgroup$
– Peter LeFanu Lumsdaine
May 27 at 10:29
|
show 14 more comments
$begingroup$
For all my homework in real analysis, when I've been asked to show that a function is continuous, I just found a single $x_n in D$ and showed that when $x_n rightarrow x_0$, $f(x_n) rightarrow f(x_0)$. Apparently, the sequence definition (as opposed to the epsilon delta definition) is (basically) only used to prove a function is not continuous, and I can't prove a function is continuous because then I'd have to show this is true for all possible sequences? Am I doing the math wrongly? Should I always use the epsilon delta definition when trying to prove that a function is continuous?
real-analysis continuity
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
$endgroup$
For all my homework in real analysis, when I've been asked to show that a function is continuous, I just found a single $x_n in D$ and showed that when $x_n rightarrow x_0$, $f(x_n) rightarrow f(x_0)$. Apparently, the sequence definition (as opposed to the epsilon delta definition) is (basically) only used to prove a function is not continuous, and I can't prove a function is continuous because then I'd have to show this is true for all possible sequences? Am I doing the math wrongly? Should I always use the epsilon delta definition when trying to prove that a function is continuous?
real-analysis continuity
real-analysis continuity
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
edited May 29 at 9:53
glS
1,2451 gold badge9 silver badges24 bronze badges
1,2451 gold badge9 silver badges24 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
asked May 27 at 0:45
MinYoung KimMinYoung Kim
35611 bronze badges
35611 bronze badges
1
$begingroup$
That would be incorrect.
$endgroup$
– Math1000
May 27 at 0:53
22
$begingroup$
oh god. this is not good.
$endgroup$
– MinYoung Kim
May 27 at 0:54
4
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
$endgroup$
– Arturo Magidin
May 27 at 0:55
27
$begingroup$
Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
$endgroup$
– Arturo Magidin
May 27 at 0:55
3
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
$endgroup$
– Peter LeFanu Lumsdaine
May 27 at 10:29
|
show 14 more comments
1
$begingroup$
That would be incorrect.
$endgroup$
– Math1000
May 27 at 0:53
22
$begingroup$
oh god. this is not good.
$endgroup$
– MinYoung Kim
May 27 at 0:54
4
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
$endgroup$
– Arturo Magidin
May 27 at 0:55
27
$begingroup$
Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
$endgroup$
– Arturo Magidin
May 27 at 0:55
3
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
$endgroup$
– Peter LeFanu Lumsdaine
May 27 at 10:29
1
1
$begingroup$
That would be incorrect.
$endgroup$
– Math1000
May 27 at 0:53
$begingroup$
That would be incorrect.
$endgroup$
– Math1000
May 27 at 0:53
22
22
$begingroup$
oh god. this is not good.
$endgroup$
– MinYoung Kim
May 27 at 0:54
$begingroup$
oh god. this is not good.
$endgroup$
– MinYoung Kim
May 27 at 0:54
4
4
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
$endgroup$
– Arturo Magidin
May 27 at 0:55
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
$endgroup$
– Arturo Magidin
May 27 at 0:55
27
27
$begingroup$
Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
$endgroup$
– Arturo Magidin
May 27 at 0:55
$begingroup$
Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
$endgroup$
– Arturo Magidin
May 27 at 0:55
3
3
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
$endgroup$
– Peter LeFanu Lumsdaine
May 27 at 10:29
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
$endgroup$
– Peter LeFanu Lumsdaine
May 27 at 10:29
|
show 14 more comments
4 Answers
4
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oldest
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$begingroup$
This is indeed incorrect. Take for example the function
$$f(x) = begin{cases}
1, text{ if } xin mathbb{Q}\
0, text{ otherwise}
end{cases}$$
This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
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3
$begingroup$
I am baffled by the popularity of my answer.
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– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
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– MinYoung Kim
May 31 at 4:20
add a comment
|
$begingroup$
Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_nto x$ then $f(x_n) to f(x)$
One specific sequence does not prove continuity.
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
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What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D rightarrow mathbb{R}$ is continuous, you need to say "Let $x_0 in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.
From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$lim x_n = x_0$" to "$lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+frac{1}{n})_n$.
You also asked if you should always use $epsilon-delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $epsilon-delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $epsilon=1$ and showing there exists some $delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x in D$ and $|x-x_0|<delta$. Like...congrats. You did it for $epsilon=1$, but you didn't do it for $epsilon=2$. You didn't do it for $epsilon>0$. It's just for $epsilon-delta$ definitions, the idea of using arbitrary $epsilon$'s and $delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
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It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
add a comment
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Is not enought find a particular sequence $x_{n}$ such that $x_{n} to x_{0} implies f(x_{n}) to f(x_{0})$
Well, $f:[0,2] to mathbb{R}$ such that $f(x) = 0$ if $ xin [0,1] $ and $f(x) = 1 $ if $x in (1,2]$ holds if we take $x_{n} = 1-frac{1}{n}$ then $x_{n} to 1$ and $ f(1-frac{1}{n}) to f(1)$ but $f$ is not continuos in $x = 1$
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is indeed incorrect. Take for example the function
$$f(x) = begin{cases}
1, text{ if } xin mathbb{Q}\
0, text{ otherwise}
end{cases}$$
This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
$endgroup$
3
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
add a comment
|
$begingroup$
This is indeed incorrect. Take for example the function
$$f(x) = begin{cases}
1, text{ if } xin mathbb{Q}\
0, text{ otherwise}
end{cases}$$
This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
$endgroup$
3
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
add a comment
|
$begingroup$
This is indeed incorrect. Take for example the function
$$f(x) = begin{cases}
1, text{ if } xin mathbb{Q}\
0, text{ otherwise}
end{cases}$$
This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
$endgroup$
This is indeed incorrect. Take for example the function
$$f(x) = begin{cases}
1, text{ if } xin mathbb{Q}\
0, text{ otherwise}
end{cases}$$
This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
answered May 27 at 0:57
tiatia
1,0153 silver badges11 bronze badges
1,0153 silver badges11 bronze badges
3
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
add a comment
|
3
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
3
3
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I am baffled by the popularity of my answer.
$endgroup$
– tia
May 30 at 11:41
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
$begingroup$
I just followed the crowd
$endgroup$
– MinYoung Kim
May 31 at 4:20
add a comment
|
$begingroup$
Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_nto x$ then $f(x_n) to f(x)$
One specific sequence does not prove continuity.
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_nto x$ then $f(x_n) to f(x)$
One specific sequence does not prove continuity.
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_nto x$ then $f(x_n) to f(x)$
One specific sequence does not prove continuity.
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
$endgroup$
Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_nto x$ then $f(x_n) to f(x)$
One specific sequence does not prove continuity.
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
edited May 27 at 1:04
Ivo Terek
50.6k9 gold badges58 silver badges158 bronze badges
50.6k9 gold badges58 silver badges158 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
answered May 27 at 0:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
53.6k4 gold badges27 silver badges73 bronze badges
53.6k4 gold badges27 silver badges73 bronze badges
add a comment
|
add a comment
|
$begingroup$
What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D rightarrow mathbb{R}$ is continuous, you need to say "Let $x_0 in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.
From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$lim x_n = x_0$" to "$lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+frac{1}{n})_n$.
You also asked if you should always use $epsilon-delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $epsilon-delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $epsilon=1$ and showing there exists some $delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x in D$ and $|x-x_0|<delta$. Like...congrats. You did it for $epsilon=1$, but you didn't do it for $epsilon=2$. You didn't do it for $epsilon>0$. It's just for $epsilon-delta$ definitions, the idea of using arbitrary $epsilon$'s and $delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
$endgroup$
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
add a comment
|
$begingroup$
What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D rightarrow mathbb{R}$ is continuous, you need to say "Let $x_0 in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.
From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$lim x_n = x_0$" to "$lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+frac{1}{n})_n$.
You also asked if you should always use $epsilon-delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $epsilon-delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $epsilon=1$ and showing there exists some $delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x in D$ and $|x-x_0|<delta$. Like...congrats. You did it for $epsilon=1$, but you didn't do it for $epsilon=2$. You didn't do it for $epsilon>0$. It's just for $epsilon-delta$ definitions, the idea of using arbitrary $epsilon$'s and $delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
$endgroup$
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
add a comment
|
$begingroup$
What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D rightarrow mathbb{R}$ is continuous, you need to say "Let $x_0 in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.
From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$lim x_n = x_0$" to "$lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+frac{1}{n})_n$.
You also asked if you should always use $epsilon-delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $epsilon-delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $epsilon=1$ and showing there exists some $delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x in D$ and $|x-x_0|<delta$. Like...congrats. You did it for $epsilon=1$, but you didn't do it for $epsilon=2$. You didn't do it for $epsilon>0$. It's just for $epsilon-delta$ definitions, the idea of using arbitrary $epsilon$'s and $delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
$endgroup$
What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D rightarrow mathbb{R}$ is continuous, you need to say "Let $x_0 in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.
From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$lim x_n = x_0$" to "$lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+frac{1}{n})_n$.
You also asked if you should always use $epsilon-delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $epsilon-delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $epsilon=1$ and showing there exists some $delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x in D$ and $|x-x_0|<delta$. Like...congrats. You did it for $epsilon=1$, but you didn't do it for $epsilon=2$. You didn't do it for $epsilon>0$. It's just for $epsilon-delta$ definitions, the idea of using arbitrary $epsilon$'s and $delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
answered May 27 at 1:08
Spencer KraislerSpencer Kraisler
8934 silver badges13 bronze badges
8934 silver badges13 bronze badges
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
add a comment
|
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
$begingroup$
It may be helpful to use the term universal generalization, which is the name of this rule of inference.
$endgroup$
– Daniel R. Collins
May 27 at 17:30
add a comment
|
$begingroup$
Is not enought find a particular sequence $x_{n}$ such that $x_{n} to x_{0} implies f(x_{n}) to f(x_{0})$
Well, $f:[0,2] to mathbb{R}$ such that $f(x) = 0$ if $ xin [0,1] $ and $f(x) = 1 $ if $x in (1,2]$ holds if we take $x_{n} = 1-frac{1}{n}$ then $x_{n} to 1$ and $ f(1-frac{1}{n}) to f(1)$ but $f$ is not continuos in $x = 1$
answered May 27 at 0:55
ZAFZAF
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$endgroup$
add a comment
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$begingroup$
Is not enought find a particular sequence $x_{n}$ such that $x_{n} to x_{0} implies f(x_{n}) to f(x_{0})$
Well, $f:[0,2] to mathbb{R}$ such that $f(x) = 0$ if $ xin [0,1] $ and $f(x) = 1 $ if $x in (1,2]$ holds if we take $x_{n} = 1-frac{1}{n}$ then $x_{n} to 1$ and $ f(1-frac{1}{n}) to f(1)$ but $f$ is not continuos in $x = 1$
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
$endgroup$
add a comment
|
$begingroup$
Is not enought find a particular sequence $x_{n}$ such that $x_{n} to x_{0} implies f(x_{n}) to f(x_{0})$
Well, $f:[0,2] to mathbb{R}$ such that $f(x) = 0$ if $ xin [0,1] $ and $f(x) = 1 $ if $x in (1,2]$ holds if we take $x_{n} = 1-frac{1}{n}$ then $x_{n} to 1$ and $ f(1-frac{1}{n}) to f(1)$ but $f$ is not continuos in $x = 1$
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
$endgroup$
Is not enought find a particular sequence $x_{n}$ such that $x_{n} to x_{0} implies f(x_{n}) to f(x_{0})$
Well, $f:[0,2] to mathbb{R}$ such that $f(x) = 0$ if $ xin [0,1] $ and $f(x) = 1 $ if $x in (1,2]$ holds if we take $x_{n} = 1-frac{1}{n}$ then $x_{n} to 1$ and $ f(1-frac{1}{n}) to f(1)$ but $f$ is not continuos in $x = 1$
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
answered May 27 at 0:55
ZAFZAF
1,0408 bronze badges
1,0408 bronze badges
add a comment
|
add a comment
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$begingroup$
That would be incorrect.
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– Math1000
May 27 at 0:53
22
$begingroup$
oh god. this is not good.
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– MinYoung Kim
May 27 at 0:54
4
$begingroup$
The sequential defintion says that for all sequences $(x_n)$ such that $x_nto x$, we have $f(x_n)to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_nto x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire.
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– Arturo Magidin
May 27 at 0:55
27
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Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.”
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– Arturo Magidin
May 27 at 0:55
3
$begingroup$
@MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n to x_0$”.
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– Peter LeFanu Lumsdaine
May 27 at 10:29