# Changing the order of quantifiers in the definition of continuity of function

I wonder what if we could change the order of the quantifiers in the definition of continuity of function. I mean

$$a)$$ For any number $$\forall \delta >0,$$ there exists some number $$\exists \varepsilon =\varepsilon \left( \delta ,x_{ 0 } \right) \quad$$ such that $$\quad \left| x-{ x }_{ 0 } \right| <\delta \quad \Rightarrow \left| f\left( x \right)-f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad$$

$$b)$$ For any number $$\forall \varepsilon >0,$$ there exists some number $$\exists \delta =\delta \left( \varepsilon ,x_{ 0 } \right) \quad$$ such that $$\quad \left| f\left( x \right) -f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad \Rightarrow \quad \left| x-{ x }_{ 0 } \right| <\delta \quad$$

$$c)$$ For any number $$\forall \delta >0,$$ there exists some number $$\exists \varepsilon =\varepsilon \left( \delta ,x_{ 0 } \right) \quad$$ such that $$\quad \left| f\left( x \right) -f\left( { x }_{ 0 } \right) \right| <\varepsilon \quad \quad \Rightarrow \quad \left| x-{ x }_{ 0 } \right| <\delta$$

I know in these variants function is not continuous at $$x=x_0$$ point, but I can't prove it, or can't get a really good counterexample.

• What are the variables $i$ and $f$? Commented Jul 13, 2019 at 13:22
• For (a) consider the simple counter example f(x)= 1 if x is not 0, f(0)= 0. Given any $\delta> 0$, let $\epsilon= 2$. |f(x)- f(y)| for any x and y is either 0 or 1 both of which are less than 2. Commented Jul 13, 2019 at 13:23
• @Bernard,it is the word "if" Commented Jul 13, 2019 at 13:24
• @user247327 would be more interesting to see what functions satisfy this. Commented Jul 13, 2019 at 13:24
• Then write it as a word… B.t.w., don't you mean ‘such that’? Commented Jul 13, 2019 at 13:25

a) This is satisfied if e.g. function $$f$$ is bounded.
b) Let $$f(x)=g(x)x$$ where $$g$$ is any function that takes values in $$\mathbb R\setminus[-1,1]$$.
Then $$|f(x)-f(x_0)|\geq |x-x_0|$$ so that $$\delta=\varepsilon$$ works.
c) Let $$f$$ be a function that satisfies $$x\neq x_0\implies |f(x)-f(x_0)|>1$$.
Then $$\varepsilon=1$$ works.
In all cases $$f$$ can be chosen to be not continuous at $$x_0$$.