Question about the proof of the $\epsilon-\delta$ definition of continuity

Question

I am currently trying to get my head around the proof of the definition of continuity of a function given in my Elementary Analysis textbook. The definition given is: Let $f$ be a real-valued function whose domain is a subset of $\mathbb{R}$.Then $f$ is continuous at $x_0$ in $dom(f)$ if and only if: for each $\epsilon>0$, there exists $\delta>0$ such that $x\in dom(f)$ and $|{x-x_0}|$ imply $|f(x)-f(x_0)|<\epsilon $.

The proof goes as follows:

Assume $f$ is continuous at $x_0$ but the definition fails. Then there exists $\epsilon>0$ so that the implication "$x\in dom(f)$ and $|{x-x_0}|$ imply $|f(x)-f(x_0)|<\epsilon $" fails for each $\delta>0$.

In particular, the implication "$x\in dom(f)$ and $|{x-x_0}|< \frac{1}{n}$ imply $|f(x)-f(x_0)|<\epsilon $" fails for each $n\in\mathbb{N}$.

Could somebody please explain to me why the implication "$x\in dom(f)$ and $|{x-x_0}|< \frac{1}{n}$ imply $|f(x)-f(x_0)|<\epsilon $" would be true when the definition holds, i.e. why has $\frac{1}{n}$ been used rather than $\delta$ as surely it could be the case that $\delta$ is a natural number? Thank you in advance.

Answer 1

In general, when we say that some statement "fails for every $\delta > 0,$" then the statement must "fail" when $\delta = 1,$ because $1 > 0.$ If the statement somehow did not fail when $\delta = 1,$ then $\delta = 1$ would be a counterexample for the claim that the statement "fails for every $\delta > 0,$" and the claim would be false.

Likewise, since $\frac12 > 0,$ the statement also must fail when $\delta = \frac12.$ Similar reasons show it fails also when $\delta = \frac13,$ when $\delta = \frac14,$ when $\delta = \frac15,$ and indeed when $\delta = \frac1n$ where $n$ is any positive integer.

Of course the statement also fails when $\delta = 2,$ when $\delta = 3,$ when $\delta = 17,$ when $\delta = \sqrt2,$ and when $\delta = \pi.$ All of that is true, but it does not negate any of the claims about the statement failing for any of the values $\delta = \frac1n$ for any integer $n.$ All of those claims are true, provided that the statement fails for all $\delta > 0.$

It often happens in mathematics that we take something that is said to be true for a very large number of cases, and focus our attention on only a small subset of those cases, because those cases are enough to show whatever it is we need to show next.

To see whether the proof is correct, we have to look at what is done with the limited set of cases afterwards, that is, what is the rest of the proof. The part you have copied for us is (presumably) barely the beginning.

Answer 2

I am not sure how you can "prove" a definition. You must be trying to show that the $\epsilon - \delta$ definition of continuity is equivalent to some $other$ definition of continuity. You have not specified what your original definition of continuity is. Do you already have the $\epsilon - \delta$ definition of limit of a function at a point? If so, then I would assume your original definition of continuity is something like the following.

A real valued function $f$ is continuous at a point $x_0 \in \text{dom}(f)$ if and only if $\lim_{x\rightarrow x_0}f(x) = f(x_0)$.

If this is your original definition and you already have the $\epsilon - \delta$ definition of function limits, then here is how you could prove that if $f$ is continuous at $x_0$ then for all $\epsilon > 0$ there exists a $\delta > 0$ such that $x\in \text{dom}(f)$ and $|x - x_0| < \delta$ implies that $|f(x) - f(x_0)| < \epsilon$. (Note that you would still have to prove the other direction to get the full equivalence result)

Suppose $f$ is continuous at a point $x_0 \in\text{dom}(f)$. Then $\lim_{x\rightarrow x_0}f(x) = f(x_0)$. let $\epsilon > 0$. Then using the definition of the limit, there exists a $\delta >0$ such that $x\in \text{dom}(f)$ and $0 < |x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \epsilon$. Thus, we only need to make sure $|f(x) - f(x_0)| < \epsilon$ when $|x - x_0| = 0$. So, suppose $|x - x_0| = 0$. Then we have $x - x_0 = 0$, so $x = x_0$. But then $f(x) = f(x_0)$, so $|f(x) - f(x_0)| = 0 < \epsilon$. Thus, we have shown that $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \epsilon$. Therefore, if $f$ is continuous at $x_0 \in \text{dom}(f)$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $x \in \text{dom}(f)$ and $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \epsilon$. That is, the $\epsilon - \delta$ definition of continuity follows from the "limit" definition of continuity.

The only difference between the $\epsilon - \delta$ definitions of the limit and continuity of a function at a point $x_0$ is that in the limit definition we consider points $x\in\text{dom}(f)$ such that $0 < |x - x_0| < \delta$ (which $excludes$ $x = x_0$), whereas in the continuity definition we consider any $x\in\text{dom}(f)$ such that $|x - x_0| < \delta$ (which $includes$ $x = x_0$).