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Does $implies wf(y)<epsilon$ , mean $|f(y)-f(x)|<epsilon?$
$begingroup$
Right here, the poster claimed that for $epsilon>0;text{and};xin I,$ $$|y-x|<deltaimplies wf(y)<epsilon$$
which implies that the $f$ is continuous at $x.$
Question: Does $implies wf(y)<epsilon$ , mean $|f(y)-f(x)|<epsilon?$
real-analysis analysis definition riemann-integration
$endgroup$
add a comment |
$begingroup$
Right here, the poster claimed that for $epsilon>0;text{and};xin I,$ $$|y-x|<deltaimplies wf(y)<epsilon$$
which implies that the $f$ is continuous at $x.$
Question: Does $implies wf(y)<epsilon$ , mean $|f(y)-f(x)|<epsilon?$
real-analysis analysis definition riemann-integration
$endgroup$
add a comment |
$begingroup$
Right here, the poster claimed that for $epsilon>0;text{and};xin I,$ $$|y-x|<deltaimplies wf(y)<epsilon$$
which implies that the $f$ is continuous at $x.$
Question: Does $implies wf(y)<epsilon$ , mean $|f(y)-f(x)|<epsilon?$
real-analysis analysis definition riemann-integration
$endgroup$
Right here, the poster claimed that for $epsilon>0;text{and};xin I,$ $$|y-x|<deltaimplies wf(y)<epsilon$$
which implies that the $f$ is continuous at $x.$
Question: Does $implies wf(y)<epsilon$ , mean $|f(y)-f(x)|<epsilon?$
real-analysis analysis definition riemann-integration
real-analysis analysis definition riemann-integration
asked Jan 24 at 15:30
user560896
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$begingroup$
The poster's proof is not even correct, because their set $I$ depends on $epsilon$. It appears they are using $omega f(X)$ to mean $sup_{xin X}f(x)-inf_{x in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' in (x-delta,x+delta) cap [a,b]$, we have $|f(y)-f(y')|<epsilon$ (the best you could put there is $2epsilon$). It would also not be correct to put $|f(x)-f(y)|<frac{epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.
answered Jan 24 at 16:04
kccukccu
10.6k11229
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The poster's proof is not even correct, because their set $I$ depends on $epsilon$. It appears they are using $omega f(X)$ to mean $sup_{xin X}f(x)-inf_{x in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' in (x-delta,x+delta) cap [a,b]$, we have $|f(y)-f(y')|<epsilon$ (the best you could put there is $2epsilon$). It would also not be correct to put $|f(x)-f(y)|<frac{epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.
answered Jan 24 at 16:04
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
The poster's proof is not even correct, because their set $I$ depends on $epsilon$. It appears they are using $omega f(X)$ to mean $sup_{xin X}f(x)-inf_{x in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' in (x-delta,x+delta) cap [a,b]$, we have $|f(y)-f(y')|<epsilon$ (the best you could put there is $2epsilon$). It would also not be correct to put $|f(x)-f(y)|<frac{epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.
answered Jan 24 at 16:04
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
The poster's proof is not even correct, because their set $I$ depends on $epsilon$. It appears they are using $omega f(X)$ to mean $sup_{xin X}f(x)-inf_{x in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' in (x-delta,x+delta) cap [a,b]$, we have $|f(y)-f(y')|<epsilon$ (the best you could put there is $2epsilon$). It would also not be correct to put $|f(x)-f(y)|<frac{epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.
answered Jan 24 at 16:04
kccukccu
10.6k11229
$endgroup$
The poster's proof is not even correct, because their set $I$ depends on $epsilon$. It appears they are using $omega f(X)$ to mean $sup_{xin X}f(x)-inf_{x in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' in (x-delta,x+delta) cap [a,b]$, we have $|f(y)-f(y')|<epsilon$ (the best you could put there is $2epsilon$). It would also not be correct to put $|f(x)-f(y)|<frac{epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.
answered Jan 24 at 16:04
kccukccu
10.6k11229
answered Jan 24 at 16:04
kccukccu
10.6k11229
answered Jan 24 at 16:04
kccukccu
10.6k11229
answered Jan 24 at 16:04
kccukccu
10.6k11229
10.6k11229
add a comment |
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