# Another statement using the $\epsilon$- $\delta$- definition

The exercise is as follows: Given a function $f : D \mapsto \mathbb R$ Are the following conditions stronger, weaker or not comparable with continuity?

$$\forall a \in D, \exists \epsilon \gt 0, \forall \delta \gt 0, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon$$

The only thing that is different from the $\epsilon$-$\delta$-definition is that the quantifiers are contrary for the $\epsilon$ and $\delta$.

Does it suffice to say that the statement is weaker than continuity because the $\exists$-quantifier is weaker than the $\forall$?

I can't come up with better arguments for it.

That was what we were taught in class but it's so self-explanatory that it shouldn't even be in our exercise lessons.

Another statement is the following:

$$\forall \epsilon \gt 0, \exists \delta \gt 0,\forall a \in D, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon$$

It is a stronger statement and the argument that was used here is :

$Q(a,\delta) \Rightarrow \forall a \qquad \exists \delta \Rightarrow Q(\mu, \delta)$

I don't remember what the $\mu$ stands for? I think the Q is just a symbol to say it's a statement.

It is not comparable. A function that is continuous on D need not satisfy that statement, and a function that satisfies that statement need not be continuous.

In general, a statement $A$ is considered to be stronger than $B$ if the statement $A \to B$ is true, but $B \to A$ is false.

As an example, differentiability is stronger than continuity. If a function is differentiable at a point, then it is continuous at that point, but the reverse is not necessarily true.

EDIT- In response to your comment, I don't know what to make of the argument that was used. I don't know what $\mu$ is intended to mean either. It is stronger because a function that satisfies the 2nd statement will necessarily be continuous, but a continuous function need not satisfy the 2nd statement. The 2nd statement is actually the definition of "uniform continuity". As an interesting side note I will mention that when D is a closed bounded interval, the 2nd statement is equivalent to continuity, but in general for arbitrary D uniform continuity will be stronger (for the reasons mentioned above).

• You need to compare the two statements presented here with the definition of continuity (So, $\forall a \in D \forall \epsilon \gt 0 \exists \delta \gt 0 \forall x \in D |x-a|\lt \delta \Rightarrow |f(x) -f(a)|\lt \epsilon$ . Not compare these two statements with eachother. – Anonymous196 Jan 17 '18 at 21:05
• O weird because my Professor assistant told us it was weaker or it was my mistake. – Anonymous196 Jan 17 '18 at 21:08
• Thx for the example it makes more sense now. – Anonymous196 Jan 17 '18 at 21:09
• And do you know the purpose of using the argument for the 2nd statement? – Anonymous196 Jan 17 '18 at 21:15
• @AnonymousI Yes and No. See edit – David Reed Jan 17 '18 at 21:25

The first statement:

$$\forall a \in D, \exists \varepsilon > 0, \forall \delta > 0, \forall x \in D: |x-a| < \delta \implies |f(x) - f(a)| < \varepsilon$$

If you think about it hard enough, all it is saying is that the function is bounded.

So, consider the function $f(x) = \begin{cases} 0 & x < 0 \\ 1 & x \ge 0 \end{cases}$. One can verify that $f$ satisfies the first statement (by letting $\varepsilon = 2$ for every $a \in D$). However, it is not continuous.

Consider the function $g(x) = x$. It is continuous, but does not satisfy the first statement.

Therefore, the first statement is not comparable with continuity.

The second statement:

$$\forall \varepsilon > 0, \exists \delta > 0,\forall a \in D, \forall x \in D: |x-a| < \delta \implies |f(x) - f(a)| < \varepsilon$$

This statement is the definition of uniform continuity. It is a standard exercise to prove that every uniformly continuous function is continuous, but that there are continuous functions which are not uniformly continuous, so the second statement is stronger than continuity.

• That's very good explained with easy to understand examples. Thank you. – Anonymous196 Jan 17 '18 at 22:20
• +1. I think this question was on the Math GRE a few years ago – qbert Jan 17 '18 at 22:35