If a function is continuous everywhere, but undefined at one point, is it still continuous?
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This is a question regarding the definition of continuity.
My understanding of continuity is that a function is continuous at a point when it holds that $$lim_{xto a^-}f(x) = f(a) = lim_{xto a^+}f(x) quad quad (1)$$
The book I'm currently reading has this image:
Note here that $f(x)$ is defined for $x=3$, but $g(x)$ is not.
This is followed by text stating that
g(x) is continuous because $D_g = [0, 6]text{\}{3}$, thus it is continuous for all values in its domain.
My point of contention here is that, how can we say that it is continuous at $x=3$ when $g(3)$ does not exist? Referring to the aforementioned definition $(1)$ that the limits converge to the actual value at this point.
I would have immediately declared both cases as jump discontinuities.
Am I mistaken here? Does $g(x)$ illustrate an exception to $(1)$?
limits continuity piecewise-continuity
asked Nov 20 at 9:58
Alec
2,18211538
add a comment |
up vote
6
down vote
favorite
This is a question regarding the definition of continuity.
My understanding of continuity is that a function is continuous at a point when it holds that $$lim_{xto a^-}f(x) = f(a) = lim_{xto a^+}f(x) quad quad (1)$$
The book I'm currently reading has this image:
Note here that $f(x)$ is defined for $x=3$, but $g(x)$ is not.
This is followed by text stating that
g(x) is continuous because $D_g = [0, 6]text{\}{3}$, thus it is continuous for all values in its domain.
My point of contention here is that, how can we say that it is continuous at $x=3$ when $g(3)$ does not exist? Referring to the aforementioned definition $(1)$ that the limits converge to the actual value at this point.
I would have immediately declared both cases as jump discontinuities.
Am I mistaken here? Does $g(x)$ illustrate an exception to $(1)$?
limits continuity piecewise-continuity
asked Nov 20 at 9:58
Alec
2,18211538
7
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
1
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
This is a question regarding the definition of continuity.
My understanding of continuity is that a function is continuous at a point when it holds that $$lim_{xto a^-}f(x) = f(a) = lim_{xto a^+}f(x) quad quad (1)$$
The book I'm currently reading has this image:
Note here that $f(x)$ is defined for $x=3$, but $g(x)$ is not.
This is followed by text stating that
g(x) is continuous because $D_g = [0, 6]text{\}{3}$, thus it is continuous for all values in its domain.
My point of contention here is that, how can we say that it is continuous at $x=3$ when $g(3)$ does not exist? Referring to the aforementioned definition $(1)$ that the limits converge to the actual value at this point.
I would have immediately declared both cases as jump discontinuities.
Am I mistaken here? Does $g(x)$ illustrate an exception to $(1)$?
limits continuity piecewise-continuity
asked Nov 20 at 9:58
Alec
2,18211538
This is a question regarding the definition of continuity.
My understanding of continuity is that a function is continuous at a point when it holds that $$lim_{xto a^-}f(x) = f(a) = lim_{xto a^+}f(x) quad quad (1)$$
The book I'm currently reading has this image:
Note here that $f(x)$ is defined for $x=3$, but $g(x)$ is not.
This is followed by text stating that
g(x) is continuous because $D_g = [0, 6]text{\}{3}$, thus it is continuous for all values in its domain.
My point of contention here is that, how can we say that it is continuous at $x=3$ when $g(3)$ does not exist? Referring to the aforementioned definition $(1)$ that the limits converge to the actual value at this point.
I would have immediately declared both cases as jump discontinuities.
Am I mistaken here? Does $g(x)$ illustrate an exception to $(1)$?
limits continuity piecewise-continuity
limits continuity piecewise-continuity
asked Nov 20 at 9:58
Alec
2,18211538
asked Nov 20 at 9:58
Alec
2,18211538
asked Nov 20 at 9:58
Alec
2,18211538
asked Nov 20 at 9:58
Alec
2,18211538
asked Nov 20 at 9:58
Alec
2,18211538
2,18211538
7
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
1
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44
add a comment |
7
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
1
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44
7
7
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
1
1
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44
add a comment |
2 Answers
2
active
oldest
votes
up vote
17
down vote
accepted
$G$ is continuous on the domain $[0,3)cup(3,6]$.
Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.
3 is not in the domain. For every point in the domain of $g$, we have the required convergence.
answered Nov 20 at 10:00
user3482749
1,876411
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
|
show 1 more comment
up vote
2
down vote
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.
answered Nov 20 at 10:01
Eevee Trainer
1,474216
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
$G$ is continuous on the domain $[0,3)cup(3,6]$.
Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.
3 is not in the domain. For every point in the domain of $g$, we have the required convergence.
answered Nov 20 at 10:00
user3482749
1,876411
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
|
show 1 more comment
up vote
17
down vote
accepted
$G$ is continuous on the domain $[0,3)cup(3,6]$.
Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.
3 is not in the domain. For every point in the domain of $g$, we have the required convergence.
answered Nov 20 at 10:00
user3482749
1,876411
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
|
show 1 more comment
up vote
17
down vote
accepted
up vote
17
down vote
accepted
$G$ is continuous on the domain $[0,3)cup(3,6]$.
Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.
3 is not in the domain. For every point in the domain of $g$, we have the required convergence.
answered Nov 20 at 10:00
user3482749
1,876411
$G$ is continuous on the domain $[0,3)cup(3,6]$.
Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.
3 is not in the domain. For every point in the domain of $g$, we have the required convergence.
answered Nov 20 at 10:00
user3482749
1,876411
edited Nov 20 at 10:03
answered Nov 20 at 10:00
user3482749
1,876411
answered Nov 20 at 10:00
user3482749
1,876411
answered Nov 20 at 10:00
user3482749
1,876411
1,876411
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
|
show 1 more comment
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
Damnit, I failed at proof reading.
– user3482749
Nov 20 at 10:03
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It's understandable, I've made similar derps before, myself (one earlier today in fact >_>)
– Eevee Trainer
Nov 20 at 10:05
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
It seems so obvious to me now. Why would I even consider the point at which the function is not defined? So why would that point have any impact on the continuity of the function?
– Alec
Nov 20 at 14:37
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
Proofreading or proof reading? The former is checking the text for correctness, the latter refers to reading a proof (not a good failure for a mathematician).
– Barmar
Nov 20 at 16:36
1
1
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
@Alec, don't chastize yourself. I'd say many make this "visual" error; the example is maximally counter-intuitive to drive home its point after the reader has understood the issue (i.e., it is designed to destroy the naive intuition that "continuous" means that you can draw everything with a pen without lifting it..., which was literally the motivation for "continuous" at least in school (for pupils 15'ish years old, some decades ago) in my neck of the woods).
– AnoE
Nov 20 at 16:39
|
show 1 more comment
up vote
2
down vote
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.
answered Nov 20 at 10:01
Eevee Trainer
1,474216
add a comment |
up vote
2
down vote
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.
answered Nov 20 at 10:01
Eevee Trainer
1,474216
add a comment |
up vote
2
down vote
up vote
2
down vote
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.
answered Nov 20 at 10:01
Eevee Trainer
1,474216
The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.
answered Nov 20 at 10:01
Eevee Trainer
1,474216
answered Nov 20 at 10:01
Eevee Trainer
1,474216
answered Nov 20 at 10:01
Eevee Trainer
1,474216
answered Nov 20 at 10:01
Eevee Trainer
1,474216
1,474216
add a comment |
add a comment |
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7
It is nowhere said that $g$ is continuous at $x=3$.
– Martin R
Nov 20 at 10:01
1
There is a theorem saying that $g:Dtocdots$ is continuous iff $g$ is continuous at every $xin D$. Here $3notin D$ so is irrelevant if it comes to the question whether $g$ is continuous. See here for a related question.
– drhab
Nov 20 at 10:06
Also potentially relevant: If there are finitely many points where there are discontinuities, then $g$ is continuous almost everywhere.
– Spitemaster
Nov 20 at 19:06
If a function is continuous everywhere, but undefined at one point, the sky is purple.
– immibis
Nov 20 at 21:44