Replace the epsilon and delta in the definition of continuity, what kind of continuity do we get? ...
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Replace the epsilon and delta in the definition of continuity, what kind of continuity do we get?
The 2019 Stack Overflow Developer Survey Results Are InThe epsilon-delta definition of continuityThe epsilon-delta definition of continuity$delta$ and $epsilon$ in the continuity definitionEpsilon-delta definition of continuityWhy epsilon and delta can be changed in the definition of continuity when the function is limited in an intervalCan limits be defined in a more algebraic way, instead of using the completely analytic $delta$-$epsilon$ definition?Test the Function for continuity using the Epsilon-Delta definition.Show continuity of a function using epsilon delta definitionHow to find $delta$ in the $varepsilon-delta$ definition of continuityDelta Epsilon Continuity proof exampleProof of equivalence between open-set and $varepsilon$-$delta$ definition of continuity.
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1) $f : X to Y$ is a map. For all $x_0in X$, if for every $delta > 0$, there is $varepsilon > 0$ such that $d_X(x_0, x) < delta$ whenever $d_Y(f(x_0), f(x)) < varepsilon$.
2)$f : X to Y$ is a map. For all $x_0in X$, if for every $ varepsilon> 0$, there is $delta > 0$ such that $d_Y(f(x_0), f(x)) < delta$ whenever $d_X(x_0, x) < varepsilon$.
Is map $f$ continuous in the two cases? If not, what kind of continuity is this?
For the second the case, I guess this definition is much stronger than continuity?
real-analysis calculus analysis functions continuity
asked Mar 21 at 6:53
High GPAHigh GPA
916422
$endgroup$
|
show 2 more comments
$begingroup$
1) $f : X to Y$ is a map. For all $x_0in X$, if for every $delta > 0$, there is $varepsilon > 0$ such that $d_X(x_0, x) < delta$ whenever $d_Y(f(x_0), f(x)) < varepsilon$.
2)$f : X to Y$ is a map. For all $x_0in X$, if for every $ varepsilon> 0$, there is $delta > 0$ such that $d_Y(f(x_0), f(x)) < delta$ whenever $d_X(x_0, x) < varepsilon$.
Is map $f$ continuous in the two cases? If not, what kind of continuity is this?
For the second the case, I guess this definition is much stronger than continuity?
real-analysis calculus analysis functions continuity
asked Mar 21 at 6:53
High GPAHigh GPA
916422
$endgroup$
1
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
1
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
1
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
1
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
1
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30
|
show 2 more comments
$begingroup$
1) $f : X to Y$ is a map. For all $x_0in X$, if for every $delta > 0$, there is $varepsilon > 0$ such that $d_X(x_0, x) < delta$ whenever $d_Y(f(x_0), f(x)) < varepsilon$.
2)$f : X to Y$ is a map. For all $x_0in X$, if for every $ varepsilon> 0$, there is $delta > 0$ such that $d_Y(f(x_0), f(x)) < delta$ whenever $d_X(x_0, x) < varepsilon$.
Is map $f$ continuous in the two cases? If not, what kind of continuity is this?
For the second the case, I guess this definition is much stronger than continuity?
real-analysis calculus analysis functions continuity
asked Mar 21 at 6:53
High GPAHigh GPA
916422
$endgroup$
1) $f : X to Y$ is a map. For all $x_0in X$, if for every $delta > 0$, there is $varepsilon > 0$ such that $d_X(x_0, x) < delta$ whenever $d_Y(f(x_0), f(x)) < varepsilon$.
2)$f : X to Y$ is a map. For all $x_0in X$, if for every $ varepsilon> 0$, there is $delta > 0$ such that $d_Y(f(x_0), f(x)) < delta$ whenever $d_X(x_0, x) < varepsilon$.
Is map $f$ continuous in the two cases? If not, what kind of continuity is this?
For the second the case, I guess this definition is much stronger than continuity?
real-analysis calculus analysis functions continuity
real-analysis calculus analysis functions continuity
asked Mar 21 at 6:53
High GPAHigh GPA
916422
asked Mar 21 at 6:53
High GPAHigh GPA
916422
edited Mar 21 at 7:26
High GPA
asked Mar 21 at 6:53
High GPAHigh GPA
916422
asked Mar 21 at 6:53
High GPAHigh GPA
916422
asked Mar 21 at 6:53
High GPAHigh GPA
916422
916422
1
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
1
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
1
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
1
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
1
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30
|
show 2 more comments
1
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
1
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
1
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
1
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
1
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30
1
1
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
1
1
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
1
1
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
1
1
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
1
1
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30
|
show 2 more comments
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1
$begingroup$
You have shown that $f^{-1}$ is continuous at $f(x_0)$ (if it exists at all) instead of showing that $f$ is continuous at $x_0.$
$endgroup$
– Dbchatto67
Mar 21 at 6:57
1
$begingroup$
The function is bounded in the second case
$endgroup$
– vidyarthi
Mar 21 at 7:24
1
$begingroup$
@Dbchatto67 Due to my humble love of math
$endgroup$
– High GPA
Mar 21 at 7:27
1
$begingroup$
why the downvote?
$endgroup$
– vidyarthi
Mar 21 at 7:28
1
$begingroup$
see here
$endgroup$
– vidyarthi
Mar 21 at 7:30