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

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 $$\leq$$.

• This is just continuity at $x_{0}$ though. – Indrayudh Roy Dec 16 '18 at 20:45
• 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. – RandomMathDude Dec 16 '18 at 20:47
• 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 $\epsilon$) – Sar Dec 16 '18 at 20:48