# Epsilon-delta definitions, inequality strict / non-strict?

Reading some of the analysis related posts, I have a question regarding the epsilon-delta language.

What we are taught is the inequality in the definition is strict. E.g $$\forall\varepsilon >0\ \exists\delta >0: \forall x\in D\left ( |x-a|<\delta \Longrightarrow |f(x)-f(a)|<\varepsilon \right )$$ ( definition of continuity at $a\in D$). If this is satisfied, we conclude for suitable choice of $x$ the difference between $f(x)$ and $f(a)$ is strictly less than any positive number hence it must be zero.

Intuitively, it also makes sense that it's sufficient if the inequality involving $\varepsilon$ is not strict. But how does one justify that?

Let $R(\varepsilon)$ represent a definition with strict $\varepsilon$ -inequality. Let $M(\varepsilon)$ be the same definition, but let $\varepsilon$-inequality be non-strict. Then $R(\varepsilon)$ is satisfied iff $M(\varepsilon)$ is satisfied?

The question really is if $M(\varepsilon)$ is satisfied, is then $R(\varepsilon)$ satisfied? Does one simply say that since $M\left (\frac{\varepsilon}{2}\right )$, then $R(\varepsilon)$?

I can almost be sure that we can't allow the inequality involving $\delta$ to be non-strict, otherwise we could potentially permit points where $f$ tends to infinity? [On second thought, just make $\delta$ smaller]

• Strictness is irrelevant because the inequalities must hold for any $\delta$. – Yves Daoust Jun 19 '17 at 9:39
• @YvesDaoust I get the idea. If the closed $\delta$ ball around $a$ contains a singularity, then we simply take a smaller closed ball around $a$. If $f$ is continuous at $a$, then there necessarely exists such a closed ball around $a$ that doesn't contain singularities. Noone explained these "philosophical" problems in our analysis courses. It was always strict inequalities. I was merely wondering if that made any difference. – Alvin Lepik Jun 19 '17 at 9:47

You're not correct, because we have the statement for all epsilon and delta, the strictness does not matter at all. The following are equivalent:

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

b) $$\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|\leq \delta \Rightarrow |f(x)-f(a)| \leq \varepsilon)$$

c) $$\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|< \delta \Rightarrow |f(x)-f(a)| \leq \varepsilon)$$

d) $$\forall \varepsilon>0 \exists \delta>0:\forall x \in D: (|x-a|\leq \delta \Rightarrow |f(x)-f(a)| < \varepsilon)$$

Let us prove a) $$\Leftrightarrow$$ b) and then you will understand the rest.

$$a) \Rightarrow b)$$: Given $$\varepsilon >0$$, by a) there is a $$\tilde{\delta} > 0$$ such that $$\forall x \in D: (|x-a|<\tilde{\delta} \Rightarrow |f(x)-f(a)| < \varepsilon)$$. Now choose $$\delta = \frac{1}{2} \tilde{\delta}$$, then $$\forall x \in D: (|x-a|\leq\delta<2\delta \Rightarrow |f(x)-f(a)| < \varepsilon \Rightarrow |f(x)-f(a)| \leq \varepsilon)$$.

$$b) \Rightarrow a)$$: Given $$\varepsilon >0$$, we define $$\tilde{\varepsilon} = \frac{1}{2}\varepsilon$$. By b) there is a $$\delta > 0$$ such that $$\forall x \in D: (|x-a|\leq\delta \Rightarrow |f(x)-f(a)| \leq \tilde{\varepsilon})$$.

Thus we have: $$\forall x \in D: (|x-a|<\delta \Rightarrow |x-a|\leq\delta \Rightarrow |f(x)-f(a)| \leq \tilde{\varepsilon} < \epsilon)$$. Q.E.D.