Quick epsilon-delta question

Question

I have a particular epsilon delta problem but the context does not really matter.

I had to prove that a function $f: \mathbb{R} \to \mathbb{R}$ was continuous in a point $0 \neq x_0 \in \mathbb{R}$, and at the end of the proof, I had proven:

$$\forall \epsilon >0: \exists \delta > 0: \forall x \in \mathbb{R}: |x-x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon/|x_0|$$

Is this proof correct? I think it is, since the $\epsilon/x_0$ is just epsilon multiplied by a fixed constant (I fixed $x_0$ at the beginning of the proof). It's just that I'm not entirely sure if this is allowed.

Thanks for any insights.

Answer 1

For me the question is how did you find such $\delta$? In doing so you must have said, in effect, "Here is a delta so that $\delta$ is related $\epsilon$ and $|x-x_0|< d\implies |f(x) -f(x_0)|< \frac {\epsilon}{|x_0|}$". It seems to me there has to be a way you can say "Here is a delta so that $\delta$ is related to $|x_0|\epsilon$ and $|x-x_0|< d\implies |f(x) -f(x_0)|< \frac {|x_0|\epsilon}{|x_0|}= \epsilon$".

But for that, I'd need to know exactly how you came up with the delta in the first place.

but regardless, it's allowed because you can always say:

For all $\epsilon > 0$ there is an $\epsilon' = \epsilon*{x_0}$. And for any such $\epsilon'$ there is a $\delta_{\epsilon'}$ so that whenever $|x - x_0| < \delta$ we have $|f(x) - f(x)| < \frac {\epsilon'}{|x_0|} = \epsilon$.

....or .....

For all $\epsilon > 0$ there is a $\Delta_{\epsilon}$ so that $|x-x_0| < \Delta_{\epsilon} \implies |f(x) -f(x_0)| < \frac {\epsilon}{|x_0}$. If we let $\delta = \Delta_{|x_0|\epsilon}$ be the value that would hold true so that $|x - x_0|< \Delta_{|x_0|\epsilon}=\delta \implies |f(x)-f(x_0)| < \frac {|x_0|\epsilon}{|x_0|}=\epsilon$ then ... we'd be done.

....

Or one can simply prove a Lemma:

Lemma: If for a fixed $k> 0$, it is the case that for any $\epsilon$ there exist a $\delta$ so that $|x-x_0| < \delta \implies |f(x) -f(x_0)| < k*\epsilon$ then $f$ is continuous at $x_0$.

Proof: For any $\epsilon > 0$ then $\frac {\epsilon}k > 0$ and there is a $\delta$ and $|x-x_0| < \delta \implies |f(x) - f(x_0)|< k*\frac {\epsilon}k = \epsilon$. So $f$ is continuous at $x_0$.

....

Actually why don't we teach this lemma early on? It's save all those damned triangle inequality proofs where we need to divide the epsilons in half to get $|a -c | \le |a-b| + |b-c| < \frac {\epsilon}2 + \frac{\epsilon}2 = \epsilon$. I mean what a pain are those! I mean, Amiright fellas, Amiright?

Answer 2

You have the components of a correct proof.

But it would looks a lot less ad-hoc if you start by setting $\epsilon' = \epsilon|x_0|$ and then do the current part of your proof with that $\epsilon'$ rather than the one the enemy gave you.

At the end where you've concluded $$ \cdots \cdots < \epsilon'/|x_0| $$ you can then note that the bound at the very end is actually the $\epsilon$ you're supposed to have there.