Is the following definition an equivalent definition of continuity of a function












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$begingroup$


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 $leq$.










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$endgroup$












  • $begingroup$
    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


















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 $leq$.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















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 $leq$.










share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











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asked Dec 16 '18 at 20:43









ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME

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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




















  • $begingroup$
    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












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