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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.

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  • $\begingroup$ What are the variables $i$ and $f$? $\endgroup$
    – Bernard
    Commented Jul 13, 2019 at 13:22
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    $\begingroup$ 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. $\endgroup$
    – user247327
    Commented Jul 13, 2019 at 13:23
  • $\begingroup$ @Bernard,it is the word "if" $\endgroup$
    – haqnatural
    Commented Jul 13, 2019 at 13:24
  • $\begingroup$ @user247327 would be more interesting to see what functions satisfy this. $\endgroup$
    – Jakobian
    Commented Jul 13, 2019 at 13:24
  • $\begingroup$ Then write it as a word… B.t.w., don't you mean ‘such that’? $\endgroup$
    – Bernard
    Commented Jul 13, 2019 at 13:25

1 Answer 1

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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 $x_0$.

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