We denote the following statement as $A$:

$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies |x - x_0| < \delta $$

The following answers are taken from "The epsilon-delta definition of continuity", answer by @RyanReich:

1) For $c \in \mathbb{R}$ consider the constant function $f(x) = c$. Given $x_0 \in \mathbb{R}$, taking $\varepsilon = 1$, note that for any $\delta > 0$ if $x = x_0 + \delta$ we have that

  • $| f(x) - f(x_0) | = | c - c | = 0 < 1 = \varepsilon$; but
  • $| x - x_0 | = |(x_0 + \delta ) - x_0 | = \delta \not< \delta$.

    Therefore the implication $| f(x) - f(x_0) | < \varepsilon \rightarrow | x - x_0 | < \delta$ does not hold. It follows that the function $f$ does not satisfy the given property.

2) Every function satisfy the given property because no matter what $\epsilon$ is, so long as we have $|f(x) - f(a)| < \epsilon$ just choose $\delta = 2|x - a|$, and then we have $|x - a| < \delta$.

I am a little bit confused. If the constant function satisfy the propert $A$ or not? According to first answer they do not satisfy but according to second answer they do. I mean that

$f$-constant$\Rightarrow$$A$ $\wedge$ $f$-constant$\Rightarrow$$\neg A$ which implies $f$-constant$\Rightarrow$ $f$-is not constant.

Where is my mistake i can not find.

  • $\begingroup$ You are right; the paragraph starting with “Finally” in the answer you refer to is wrong. $\endgroup$ – egreg Dec 6 '18 at 14:05

I think your error may be in the direction of implication

Your link says $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } |x - a| < \delta \implies |f(x) - f(a)| < \epsilon$$ but you seem to have interpreted it as $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } |f(x) - f(a)| < \epsilon \implies |x - a| < \delta$$

  • 1
    $\begingroup$ The problem here is that the answerer there maintains that the “reverse condition” is satisfied by every function. $\endgroup$ – egreg Dec 6 '18 at 14:10
  • $\begingroup$ @egreg - Yes and that was pointed out in the comments to that answer "The reverse definition does not allow $δ$ to depend upon $x$. The statement is that a single $δ$ should work for all $x$. Any function with bounded range, for example, would fail the reverse condition" and acknowledged as correct by the answerer, even if the answer was not edited $\endgroup$ – Henry Dec 6 '18 at 14:20

You are right.

Take for simplicity $f(x)=0$ and $x_0=0$. We'd like to prove that statement $A$ is false for this case.

Indeed, take $\varepsilon=1$. Then, no matter what $\delta$ is, we can find $x$ such that $|f(x)-f(0)|<1$ and $|x-0|\not<\delta$; just take $x=2\delta$.

The last paragraph in the answer you refer to is wrong.


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