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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 $\leq$.

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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

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