Is the following definition an equivalent definition of continuity of a function
0
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Let $f:(0,1) to mathbb{R}$ be a given function.
Definition: for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - f(x_0)| leq epsilon$ .
I think its equivalent to the definition of continuity because it doesn't matter if it's $lt$ or $leq$.
continuity definition
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
add a comment |
0
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function.
Definition: for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - f(x_0)| leq epsilon$ .
I think its equivalent to the definition of continuity because it doesn't matter if it's $lt$ or $leq$.
continuity definition
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48
add a comment |
0
0
0
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function.
Definition: for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - f(x_0)| leq epsilon$ .
I think its equivalent to the definition of continuity because it doesn't matter if it's $lt$ or $leq$.
continuity definition
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
Let $f:(0,1) to mathbb{R}$ be a given function.
Definition: for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - f(x_0)| leq epsilon$ .
I think its equivalent to the definition of continuity because it doesn't matter if it's $lt$ or $leq$.
continuity definition
continuity definition
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 20:43
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
1619
$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48
add a comment |
$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48
$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48
add a comment |
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$begingroup$
This is just continuity at $x_{0}$ though.
$endgroup$
– Indrayudh Roy
Dec 16 '18 at 20:45
$begingroup$
Yes, the strict vs. non-strict inequalities won't matter for this concept. If a function satisfies your definition, find a corresponding $(epsilon_0,delta_0)$ pair. Then for any $epsilon_1>epsilon_0$ you may take any $delta_1<delta_0$ in the traditional definition of continuity. The other direction is of course similar.
$endgroup$
– RandomMathDude
Dec 16 '18 at 20:47
$begingroup$
If I understand correctly, you're asking if the definition holds for weak inequality aswell. In the case of continuity it doesn`t matter, Becasue you can always take $epsilon<epsilon'$ and have a strong inequality (Since you can take any $epsilon$)
$endgroup$
– Sar
Dec 16 '18 at 20:48