A proof about $(\epsilon,\delta)$ definition of limit

Question

The limit $\lim_{x \to c}f(x) = L$ is defined for an open interval $I$ in the domain of $f$, possibly excluding $c$, as for every $\epsilon > 0$ there is a $\delta > 0$ such that for every $x \in I$ which provides $0 < |x-c| < \delta$ it is $|f(x) - L| < \epsilon$.

I want to show that if such a limit exists for the open interval $I$ then for each open interval $I^{*}$ in the domain of $f$, which contain $c$, the limit exists as well. It seems very intuitive that this is true but I could not think of a rigorous proof about it. Is it possible to give a proof for this statement?

Yes it is possible. Have you tried drawing a picture of an example of the situation? — Paul Sundheim, Aug 10 '14 at 19:02
Hint: First show that $I^*\cap I$ is an open interval containing $c$. — Thomas Andrews, Aug 10 '14 at 19:02
Just take $\delta < \min{\delta_0,\delta'}$ where $\delta_0$ is the one you pick to guarantee $\epsilon$ closeness for $f$ and $\delta' =\min{|c-a|, |c-b|}$ with $I^=(a,b)$, then you're inside $I^$ and near $L$. — Adam Hughes, Aug 10 '14 at 19:02

score 1 · Answer 1 · answered Aug 10 '14 at 19:02

1

If $$\forall x\in I,0<|x-c|<\delta\implies |f(x)-L|<\varepsilon,$$ necessarily $$\forall x\in I^*\cap I, 0<|x-c|<\delta\implies |f(x)-L|<\varepsilon$$

where $c\in I^*$.

answered Aug 10 '14 at 19:02

idm

11,824

We don't know that $I^*\subseteq I$, but it is a good start to prove if for that case first. – Thomas Andrews Aug 10 '14 at 19:03
I just changed it :-) – idm Aug 10 '14 at 19:04
You should also make sure to include $0<|x-c|<\delta$. The zero is important – Paul Sundheim Aug 10 '14 at 19:05
I did it, thanks :-) – idm Aug 10 '14 at 19:07
It's not clear what that shows, though, unless you know that $I^\cap I$ is an ope interal containing $c$. :) It still doesn't prove it for $I^$, since $I^$ might contain more* points than $I$ that are distance within $\delta$ of $x$.... – Thomas Andrews Aug 10 '14 at 19:08
I exactly thought the same, define the intersection $I \cap I^{*}$ and show that the limit definition applies for the intersection. But somehow this did not seem complete to me. – Ufuk Can Bicici Aug 10 '14 at 19:11

score 1 · Accepted Answer · answered Aug 10 '14 at 19:16

Let $I=(a_1,b_1)$ and $I^*=(a_2,b_2)$. For $c\in I\cap I^*$, let $$\delta_0=\min(|c-a_1|,|c-a_2|,|c-b_1|,|c-b_2|)$$

Then $\delta_0>0$ and if $|x-c|<\delta_0$ then $x\in I\cap I^*$.

Now, assume that $$\forall \epsilon>0: \exists \delta>0: \forall x\in I: \left(0<\left|x-c\right|<\delta \implies $|f(x)-L|<\epsilon\right)$$

Then, given $\epsilon>0$, find the $\delta$ and define $\delta^*=\min(\delta,\delta_0)$.

Now if $0<|x-c|<\delta^*$, we have $x\in I^*$ and $x\in I$ and $|x-c|<\delta_0\leq\delta$. So $|f(x)-L|<\epsilon$.

Can we interpret the last line as the following: If $\delta_0 > \delta$, then we a subset of $I \cap I^{}$ which provides the limit definition. If we have $\delta_0 \leq \delta$ then the whole $I \cap I^{}$ provides the definition. Is that right? — Ufuk Can Bicici, Aug 10 '14 at 19:45

A proof about $(\epsilon,\delta)$ definition of limit

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