Proof on the order of conditions in the $\epsilon$-$\delta$-definition

Question

Suppose $f(x)$ suffices to the following expression: $(\forall a\in D)(\exists\delta\gt0)(\forall\epsilon\gt0)(\forall x\in D)(|x-a|\lt\delta\Rightarrow|f(x)-f(a)|\lt\epsilon)$

Proof that the above expression isn't the same as $\epsilon$-$\delta$-definition for continuity.

On one hand one can start by looking for a counterexample which is a continuous function that doesn't suffice the condition and on the other hand you can look for a discontinuous function for which the expression holds.

For the $2^{nd}$ case I would use the stepfunction at a certain point a for which it intuitively holds. But I can't find a counterexample that contradicts the above expression.

Any hints?

Answer 1

Recall the definition of continuity for $f:D\to\mathbb R,D\subseteq\mathbb R$

$$(\forall a\in D)(\forall\varepsilon > 0)(\exists\delta >0)(\forall x\in D)\left (|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\varepsilon\right )$$

The condition you gave is:

$$(\forall a\in D)(\exists\delta\gt0)(\forall\epsilon\gt0)(\forall x\in D)(|x-a|\lt\delta\Rightarrow|f(x)-f(a)|\lt\epsilon) $$

But now we must have a $\delta$ that is universal for every $\varepsilon$ opposed to finding a $\delta = \delta (\varepsilon)$. Switching the order of the quantifiers changes the expression.

For a counter-example you may consider $f(x) = x$. Fix $a\in\mathbb R$. Does there exist a $\delta$ such that the difference between any $f(x)$ and $f(a)$, where $x\in (a-\delta, a+\delta)$, is arbitrarily small?

Suppose such a $\delta >0 $ exists. Let $\varepsilon = \frac{\delta}{2}$. Then for $x = a+\delta /2\in (a-\delta,a+\delta)$ and $$|f(x)-f(a)|\geq \varepsilon\ $$

Answer 2

Let $a\in D=\Bbb R .$ Take some $\delta >0$ such that the rest of the expression is satisfied . Then for any $x\in (-\delta+a,+\delta+a)$ we have $$\forall \epsilon >0\;(|f(x)-f(a)|<\epsilon)$$ which implies that $f(x)=f(a)$ for every $x\in (-\delta+a,+\delta+a).$

BTW. In the case $D=\Bbb R$ this will also imply that $f$ is constant, i.e. $\forall x\;(f(x)=f(0)\;)$.