Epsilon-delta definition of continuity in two dimensions The 2019 Stack Overflow Developer Survey Results Are InHow to prove differentiability implies continuity with $epsilon-delta$ definition?checking slope = $0$ at a point for a function using $epsilon $, $delta $ definition$delta$ and $epsilon$ in the continuity definitionEpsilon-delta definition of continuitySwitching the quantifier of definition for ContinuityDoes the new definition imply continuity and vice versa?changing the order of the logical symbols $(forall epsilon) (existsdelta)$ by $(exists delta)(forallepsilon) $ in limit definitionDisproving continuity using epsilon deltaShowing the continuity of a function based on the continuity of another functionQuestion about the proof of the $epsilon-delta$ definition of continuity
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Epsilon-delta definition of continuity in two dimensions
The 2019 Stack Overflow Developer Survey Results Are InHow to prove differentiability implies continuity with $epsilon-delta$ definition?checking slope = $0$ at a point for a function using $epsilon $, $delta $ definition$delta$ and $epsilon$ in the continuity definitionEpsilon-delta definition of continuitySwitching the quantifier of definition for ContinuityDoes the new definition imply continuity and vice versa?changing the order of the logical symbols $(forall epsilon) (existsdelta)$ by $(exists delta)(forallepsilon) $ in limit definitionDisproving continuity using epsilon deltaShowing the continuity of a function based on the continuity of another functionQuestion about the proof of the $epsilon-delta$ definition of continuity
$begingroup$
Usually the continuity of multi-variate function is defined through something like "$delta$-disk."
Can we define such continuity with the similar one-dimension definition? As follow:
$f:mathbb R^2tomathbb R$ is continuous at $x$ iff $foralldelta>0existsepsilon>0forall yinmathbb R^2$ we have $|x-y|<epsilon$ implies $|f(y)-f(x)|<delta$.
If this definition works, then why do we need to complicate the problem?
If not, could you please provide a counterexample?
Similarly, there are many very complicated generalization of absolute continuity in $mathbb R^n$. Why don't people just use the same definition in $mathbb R^n$ as if in $mathbb R$ if that works? If it doesn't work in $mathbb R^n$, then could you please bring a counterexample?
1d definition": $∀ε>0 ∃δ>0$ st $∑_j(y_j−x_j)<δ⟹∑_j|f(x_j)−f(y_j)|<ε$, where each $(x_j,y_j)$ is a subinterval of the domain of f. Can it be directly used in $mathbb R^n$?
real-analysis analysis functions continuity absolute-continuity
asked Mar 24 at 5:43
High GPAHigh GPA
914422
$endgroup$
|
show 2 more comments
$begingroup$
Usually the continuity of multi-variate function is defined through something like "$delta$-disk."
Can we define such continuity with the similar one-dimension definition? As follow:
$f:mathbb R^2tomathbb R$ is continuous at $x$ iff $foralldelta>0existsepsilon>0forall yinmathbb R^2$ we have $|x-y|<epsilon$ implies $|f(y)-f(x)|<delta$.
If this definition works, then why do we need to complicate the problem?
If not, could you please provide a counterexample?
Similarly, there are many very complicated generalization of absolute continuity in $mathbb R^n$. Why don't people just use the same definition in $mathbb R^n$ as if in $mathbb R$ if that works? If it doesn't work in $mathbb R^n$, then could you please bring a counterexample?
1d definition": $∀ε>0 ∃δ>0$ st $∑_j(y_j−x_j)<δ⟹∑_j|f(x_j)−f(y_j)|<ε$, where each $(x_j,y_j)$ is a subinterval of the domain of f. Can it be directly used in $mathbb R^n$?
real-analysis analysis functions continuity absolute-continuity
asked Mar 24 at 5:43
High GPAHigh GPA
914422
$endgroup$
1
$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
2
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
1
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48
|
show 2 more comments
$begingroup$
Usually the continuity of multi-variate function is defined through something like "$delta$-disk."
Can we define such continuity with the similar one-dimension definition? As follow:
$f:mathbb R^2tomathbb R$ is continuous at $x$ iff $foralldelta>0existsepsilon>0forall yinmathbb R^2$ we have $|x-y|<epsilon$ implies $|f(y)-f(x)|<delta$.
If this definition works, then why do we need to complicate the problem?
If not, could you please provide a counterexample?
Similarly, there are many very complicated generalization of absolute continuity in $mathbb R^n$. Why don't people just use the same definition in $mathbb R^n$ as if in $mathbb R$ if that works? If it doesn't work in $mathbb R^n$, then could you please bring a counterexample?
1d definition": $∀ε>0 ∃δ>0$ st $∑_j(y_j−x_j)<δ⟹∑_j|f(x_j)−f(y_j)|<ε$, where each $(x_j,y_j)$ is a subinterval of the domain of f. Can it be directly used in $mathbb R^n$?
real-analysis analysis functions continuity absolute-continuity
asked Mar 24 at 5:43
High GPAHigh GPA
914422
$endgroup$
Usually the continuity of multi-variate function is defined through something like "$delta$-disk."
Can we define such continuity with the similar one-dimension definition? As follow:
$f:mathbb R^2tomathbb R$ is continuous at $x$ iff $foralldelta>0existsepsilon>0forall yinmathbb R^2$ we have $|x-y|<epsilon$ implies $|f(y)-f(x)|<delta$.
If this definition works, then why do we need to complicate the problem?
If not, could you please provide a counterexample?
Similarly, there are many very complicated generalization of absolute continuity in $mathbb R^n$. Why don't people just use the same definition in $mathbb R^n$ as if in $mathbb R$ if that works? If it doesn't work in $mathbb R^n$, then could you please bring a counterexample?
1d definition": $∀ε>0 ∃δ>0$ st $∑_j(y_j−x_j)<δ⟹∑_j|f(x_j)−f(y_j)|<ε$, where each $(x_j,y_j)$ is a subinterval of the domain of f. Can it be directly used in $mathbb R^n$?
real-analysis analysis functions continuity absolute-continuity
real-analysis analysis functions continuity absolute-continuity
asked Mar 24 at 5:43
High GPAHigh GPA
914422
asked Mar 24 at 5:43
High GPAHigh GPA
914422
edited Mar 24 at 7:01
High GPA
asked Mar 24 at 5:43
High GPAHigh GPA
914422
asked Mar 24 at 5:43
High GPAHigh GPA
914422
asked Mar 24 at 5:43
High GPAHigh GPA
914422
914422
1
$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
2
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
1
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48
|
show 2 more comments
1
$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
2
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
1
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48
1
1
$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
2
2
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
1
1
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48
|
show 2 more comments
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$begingroup$
Given that $lvert x- y rvert < delta$ is a ball of radius $delta$ centered at $x$ (or a so-called $delta$-disk), how is the above definition different?
$endgroup$
– Gary Moon
Mar 24 at 5:50
$begingroup$
@GaryMoon I guess they are the same. Then why do we need to over-complicated it rather than saying that the definition is exactly same as the one in one dimensional case? I think math appreciates simplicity and concision.
$endgroup$
– High GPA
Mar 24 at 5:52
2
$begingroup$
I think that mentioning that to students is, in general, good and can aid in intuition/understanding. But, at the same time, you do, at some point, have to go through the process of constructing the machinery (e.g., replacing the absolute value with a higher-dimensional Euclidean norm) and actually showing that things work out like you say they will. Many things in math can be intuitive on the surface, yet take a good deal of effort to actually prove/verify. Take the Jordan curve theorem for example.
$endgroup$
– Gary Moon
Mar 24 at 6:09
$begingroup$
If there is a relation between the various dimensions you need to be careful in applying "disks" or "balls".
$endgroup$
– Moti
Mar 24 at 6:25
1
$begingroup$
Is the $1d$ definition you have in mind $forall varepsilon > 0 exists delta >0$ st $sum_j (y_j-x_j) < delta implies sum_j lvert f(x_j) - f(y_j) rvert < varepsilon$, where each $(x_j,y_j)$ is a subinterval of the domain of $f$?
$endgroup$
– Gary Moon
Mar 24 at 6:48