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so I am preparing for my final and I needed some help verifying my definitions and their justifications. So any help/feedback would be appreciated...

Let $f:(0,1) \to \mathbb{R}$ be a given function, and f is continuous. Then out of the following defnition pick the right definition, and/or equuivalent definition. If the definition is not equivalent provide explanation:

a.) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $|f(x) - f(x_0)|\lt \epsilon$.

this definition is true.

b.) 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 $|f(x) - f(x_0)|\leq \epsilon$.

this definition is equivalent.

c.) for any $\epsilon \gt 0$, for any $\delta \gt 0$ such that for all $x \in (0,1)$ and $ |x-x_0| \lt \delta$, one has $|f(x) - f(x_0)|\lt \epsilon$.

this definition is false because for any ϵ>0 there exists some δ>0 that is small enough. It can't be any delta.

d.) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $0 \lt |f(x) - f(x_0)|\leq \epsilon$.

this definition is false. I dont quite know how to provide an explanation for it tho.

these are my attempts on the identification. Could you please help me confirm these answers?

Thank you.

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d) is not true for a constant function $f$, since we then get $|f(x) - f(x_0)| = 0$. But $f$ is continuous.

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For a) a counter-example

$f(x)=\left\lbrace\begin{array}{l}0 \quad \text{if}\;x=\frac{1}{2}\\x\quad \text{if}\;x\in (0,1)\backslash\{\frac{1}{2}\}\end{array}\right.$

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