Delta-Epsilon why is delta*c less than or equal to eplison*c

Question

In a delta-epsilon proof, you find a delta that you set to epsilon. This delta is less than or equal to epsilon. Why is this? I understand that in a limit delta can shrink arbitrarily until a value as close to 0 (the distance to a) as possible but how is it that delta is less than epsilon?

Answer 1

There is no explicit ordering for $\delta$ and $\epsilon$ in an $\epsilon-\delta$ proof. The concept to understand, is that $\epsilon$ is a fixed value greater than $0$, and $\delta(\epsilon)$ such that continuity occurs.

For simplicity, consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$. Defining point continuity:

A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous at a point $x^*\in\mathbb{R}$, if for any $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that, for all $x\in \mathbb{R}$, $$\|x - x^*\| < \delta \Rightarrow \|f(x) - f(x^*)\| < \epsilon.$$

So, $\delta$ is just a positive value dependent on $\epsilon$, but there's no required ordering between $\delta$ and $\epsilon$.

Consider Lipschitz continuity of $f$, which implies continuity. For any $\epsilon > 0$, we can define $\delta(\epsilon) = \Lambda \epsilon$ and continuity of $f$ would hold, where $\Lambda\in\mathbb{R}_+$ is the Lipschitz constant. So with this, we can see that for $\Lambda \geq 1$, we could define $\delta$ such that $\delta \geq \epsilon$. But it isn't necessary to make it that large, as we just need existence.