2
$\begingroup$

I'm looking for a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that f is not continuous, and moreover $\forall \epsilon > 0, \exists \delta > 0$ such that $|f(x) - f(y)| < \epsilon \implies |x - y| < \delta$.

I can't seem to construct such a function, yet I know this should be true, given that the epsilon-delta definition of continuity has the implication the other way around. There is a similar question here but with different quantifiers.

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Let $f(x)=x$ if $x\leq 0$ and $f(x)=x+1$ if $x>0$. Then $f$ is discontinuous at $0$ but satisfies your $\epsilon$-$\delta$ condition with $\delta=\epsilon$.

$\endgroup$
1
  • $\begingroup$ Interestingly, the answer to the other question gives an example of a function much like this. $\endgroup$
    – David K
    Commented Nov 29, 2019 at 3:37
0
$\begingroup$

In some sense the function tends to finite "non - injectivity" at a point.

A couple of examples:

$1)$ Any discontinuous monotonic function

$2)$ A function $f$, $f(x)=x$ for $x\in\mathbb{Q}$, $f(x)=x+1,x\in\mathbb{R}-\mathbb{Q}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .