# Question about the post “The epsilon-delta definition of continuity”

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

Where is my mistake i can not find.

• You are right; the paragraph starting with “Finally” in the answer you refer to is wrong. – 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 $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } |f(x) - f(a)| < \epsilon \implies |x - a| < \delta$$

• The problem here is that the answerer there maintains that the “reverse condition” is satisfied by every function. – egreg Dec 6 '18 at 14:10
• @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 – 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.