# Definition and (non) equivalent definition of f being continuous

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.

## 2 Answers

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

For a) a counter-example

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